codeforces 732

A. Buy a Shovel

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for kburles. Assume that there is an unlimited number of such shovels in the shop.

In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of r burles (1 ≤ r ≤ 9).

What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of r burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.

Input

The single line of input contains two integers k and r (1 ≤ k ≤ 1000, 1 ≤ r ≤ 9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".

Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.

Output

Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.

Examples

input

117 3

output

9

input

237 7

output

1

input

15 2

output

2

#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
#define lson o<<1
#define rson o<<1|1

typedef  long long LL;
typedef unsigned long long ULL;
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('\n');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e4+120;
const int maxn=3e5+200;
const double eps=1e-12;

int main()
{
    int k,r,ans=10;
    read(k);read(r);
    for(int i=1;i<=10;i++)
    {
        if(k*i%10==0||k*i%10==r){ans=i;break;}
    }
    cout<<ans;
    return 0;
}

　　

B. Cormen — The Best Friend Of a Man

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Recently a dog was bought for Polycarp. The dog's name is Cormen. Now Polycarp has a lot of troubles. For example, Cormen likes going for a walk.

Empirically Polycarp learned that the dog needs at least k walks for any two consecutive days in order to feel good. For example, if k = 5and yesterday Polycarp went for a walk with Cormen 2 times, today he has to go for a walk at least 3 times.

Polycarp analysed all his affairs over the next n days and made a sequence of n integers a1, a2, ..., an, where ai is the number of times Polycarp will walk with the dog on the i-th day while doing all his affairs (for example, he has to go to a shop, throw out the trash, etc.).

Help Polycarp determine the minimum number of walks he needs to do additionaly in the next n days so that Cormen will feel good during all the n days. You can assume that on the day before the first day and on the day after the n-th day Polycarp will go for a walk with Cormen exactly k times.

Write a program that will find the minumum number of additional walks and the appropriate schedule — the sequence of integersb1, b2, ..., bn (bi ≥ ai), where bi means the total number of walks with the dog on the i-th day.

Input

The first line contains two integers n and k (1 ≤ n, k ≤ 500) — the number of days and the minimum number of walks with Cormen for any two consecutive days.

The second line contains integers a1, a2, ..., an (0 ≤ ai ≤ 500) — the number of walks with Cormen on the i-th day which Polycarp has already planned.

Output

In the first line print the smallest number of additional walks that Polycarp should do during the next n days so that Cormen will feel good during all days.

In the second line print n integers b1, b2, ..., bn, where bi — the total number of walks on the i-th day according to the found solutions (ai ≤ bi for all i from 1 to n). If there are multiple solutions, print any of them.

Examples

input

3 5
2 0 1

output

4
2 3 2

input

3 1
0 0 0

output

1
0 1 0

input

4 6
2 4 3 5

output

0
2 4 3 5

#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
#define lson o<<1
#define rson o<<1|1

typedef  long long LL;
typedef unsigned long long ULL;
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('\n');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e4+120;
const double eps=1e-12;
const int maxn=510;
int n,k,a[maxn];
int main()
{
    read(n);read(k);
    for(int i=1;i<=n;i++)read(a[i]);
    int ans=0;
    for(int i=2;i<=n;i++)
    {
        if(a[i-1]+a[i]<k)
        {
            ans=ans+k-a[i-1]-a[i];
            a[i]=k-a[i-1];
        }
    }
    cout<<ans<<endl;
    for(int i=1;i<=n;i++)printf("%d ",a[i]);
    return 0;
}
//贪心水题

　　

C. Sanatorium

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasiliy spent his vacation in a sanatorium, came back and found that he completely forgot details of his vacation!

Every day there was a breakfast, a dinner and a supper in a dining room of the sanatorium (of course, in this order). The only thing that Vasiliy has now is a card from the dining room contaning notes how many times he had a breakfast, a dinner and a supper (thus, the card contains three integers). Vasiliy could sometimes have missed some meal, for example, he could have had a breakfast and a supper, but a dinner, or, probably, at some days he haven't been at the dining room at all.

Vasiliy doesn't remember what was the time of the day when he arrived to sanatorium (before breakfast, before dinner, before supper or after supper), and the time when he left it (before breakfast, before dinner, before supper or after supper). So he considers any of these options. After Vasiliy arrived to the sanatorium, he was there all the time until he left. Please note, that it's possible that Vasiliy left the sanatorium on the same day he arrived.

According to the notes in the card, help Vasiliy determine the minimum number of meals in the dining room that he could have missed. We shouldn't count as missed meals on the arrival day before Vasiliy's arrival and meals on the departure day after he left.

Input

The only line contains three integers b, d and s (0 ≤ b, d, s ≤ 1018,  b + d + s ≥ 1) — the number of breakfasts, dinners and suppers which Vasiliy had during his vacation in the sanatorium.

Output

Print single integer — the minimum possible number of meals which Vasiliy could have missed during his vacation.

Examples

input

3 2 1

output

1

input

1 0 0

output

0

input

1 1 1

output

0

input

1000000000000000000 0 1000000000000000000

output

999999999999999999

#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
#define lson o<<1
#define rson o<<1|1

typedef  long long LL;
typedef unsigned long long ULL;
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('\n');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e4+120;
const double eps=1e-12;
const int maxn=510;
LL a,b,c;
int main()
{
    read(a);read(b);read(c);
    LL mmax=max(a,max(b,c));
    LL ans=0;
    if(a>=b&&a>=c)
    {
        if(a==c)
        {
            if(b==a)ans=0;
            else ans=a-b-1;
        }
        else
        {
            if(a==b)ans=a-c-1;
            else ans=a-b-1+a-c-1;
        }
    }
    else if(b>=a&&b>=c)
    {
        if(b==c)
        {
            if(a==b)ans=0;
            else ans=b-a-1;
        }
        else ans=b-a-1+b-c-1;
    }
    else if(c>=a&&c>=b)
    {
        if(c==a)
        {
            if(b==c)ans=0;
            else ans=c-b-1;
        }
        else ans=c-a-1+c-b-1;
    }
    cout<<ans<<endl;
    return 0;
}
/*
题意：给出早中晚的吃饭的次数,随意什么时候来的和走的,问最少错过了多少顿饭
思路：分情况讨论贪心啊;

*/

　　

D. Exams

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasiliy has an exam period which will continue for n days. He has to pass exams on m subjects. Subjects are numbered from 1 to m.

About every day we know exam for which one of m subjects can be passed on that day. Perhaps, some day you can't pass any exam. It is not allowed to pass more than one exam on any day.

On each day Vasiliy can either pass the exam of that day (it takes the whole day) or prepare all day for some exam or have a rest.

About each subject Vasiliy know a number ai — the number of days he should prepare to pass the exam number i. Vasiliy can switch subjects while preparing for exams, it is not necessary to prepare continuously during ai days for the exam number i. He can mix the order of preparation for exams in any way.

Your task is to determine the minimum number of days in which Vasiliy can pass all exams, or determine that it is impossible. Each exam should be passed exactly one time.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 105) — the number of days in the exam period and the number of subjects.

The second line contains n integers d1, d2, ..., dn (0 ≤ di ≤ m), where di is the number of subject, the exam of which can be passed on the day number i. If di equals 0, it is not allowed to pass any exams on the day number i.

The third line contains m positive integers a1, a2, ..., am (1 ≤ ai ≤ 105), where ai is the number of days that are needed to prepare before passing the exam on the subject i.

Output

Print one integer — the minimum number of days in which Vasiliy can pass all exams. If it is impossible, print -1.

Examples

input

7 2
0 1 0 2 1 0 2
2 1

output

5

input

10 3
0 0 1 2 3 0 2 0 1 2
1 1 4

output

9

input

5 1
1 1 1 1 1
5

output

-1

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn=1e5+10;
int n,m,a[maxn],d[maxn],p[maxn];
int check(int x)
{
    for(int i=1;i<=x;i++)if(d[i])p[d[i]]=i;
    int sum=0,cnt=0;
    for(int i=1;i<=x;i++)
    {
        if(d[i]==0){sum++;continue;}
        if(p[d[i]]!=i)sum++;
        else
        {
            if(sum<a[d[i]])return 0;
            else
            {
                cnt++;
                sum=sum-a[d[i]];
            }
        }
    }
    if(cnt==m)return 1;
    return 0;
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)scanf("%d",&d[i]);
    for(int i=1;i<=m;i++)scanf("%d",&a[i]);
    if(m>n)printf("-1\n");
    else
    {
        int l=1,r=n;
        while(l<=r)
        {
            int mid=(l+r)>>1;
            if(check(mid))r=mid-1;
            else l=mid+1;
        }
        if(r==n&&check(r)==0)printf("-1\n");
        else printf("%d\n",r+1);
    }
    return 0;
}
/*
题意：给出哪天哪课可以考试通过,和每科需要复习的天数,问最少需要多少天可以通过所有的科目;

思路：二分最少天数,然后贪心的check是否能通过所有考试
*/

　　

E. Sockets

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The ICM ACPC World Finals is coming! Unfortunately, the organizers of the competition were so busy preparing tasks that totally missed an important technical point — the organization of electricity supplement for all the participants workstations.

There are n computers for participants, the i-th of which has power equal to positive integer pi. At the same time there are m sockets available, the j-th of which has power euqal to positive integer sj. It is possible to connect the i-th computer to the j-th socket if and only if their powers are the same: pi = sj. It is allowed to connect no more than one computer to one socket. Thus, if the powers of all computers and sockets are distinct, then no computer can be connected to any of the sockets.

In order to fix the situation professor Puch Williams urgently ordered a wagon of adapters — power splitters. Each adapter has one plug and one socket with a voltage divider between them. After plugging an adapter to a socket with power x, the power on the adapter's socket becomes equal to , it means that it is equal to the socket's power divided by two with rounding up, for example  and .

Each adapter can be used only once. It is possible to connect several adapters in a chain plugging the first to a socket. For example, if two adapters are plugged one after enother to a socket with power 10, it becomes possible to connect one computer with power 3 to this socket.

The organizers should install adapters so that it will be possible to supply with electricity the maximum number of computers c at the same time. If there are several possible connection configurations, they want to find the one that uses the minimum number of adaptersu to connect c computers.

Help organizers calculate the maximum number of connected computers c and the minimum number of adapters u needed for this.

The wagon of adapters contains enough of them to do the task. It is guaranteed that it's possible to connect at least one computer.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 200 000) — the number of computers and the number of sockets.

The second line contains n integers p1, p2, ..., pn (1 ≤ pi ≤ 109) — the powers of the computers.

The third line contains m integers s1, s2, ..., sm (1 ≤ si ≤ 109) — the power of the sockets.

Output

In the first line print two numbers c and u — the maximum number of computers which can at the same time be connected to electricity and the minimum number of adapters needed to connect c computers.

In the second line print m integers a1, a2, ..., am (0 ≤ ai ≤ 109), where ai equals the number of adapters orginizers need to plug into thei-th socket. The sum of all ai should be equal to u.

In third line print n integers b1, b2, ..., bn (0 ≤ bi ≤ m), where the bj-th equals the number of the socket which the j-th computer should be connected to. bj = 0 means that the j-th computer should not be connected to any socket. All bj that are different from 0 should be distinct. The power of the j-th computer should be equal to the power of the socket bj after plugging in abj adapters. The number of non-zero bj should be equal to c.

If there are multiple answers, print any of them.

Examples

input

2 2
1 1
2 2

output

2 2
1 1
1 2

input

2 1
2 100
99

output

1 6
6
1 0

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn=2e5+10;
int n,m,a[maxn],b[maxn],num=0;
struct node
{
    int id,p;
}po[maxn],pe[maxn],ha[maxn];
int cmp(node a,node b){return a.p<b.p;}
map<int,int>mp;
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&po[i].p);
        po[i].id=i;
        mp[po[i].p]++;
    }
    for(int i=1;i<=m;i++)
    {
        scanf("%d",&pe[i].p);
        pe[i].id=i;
    }
    sort(po+1,po+n+1,cmp);
    sort(pe+1,pe+m+1,cmp);
    int c=0,u=0;
    for(int i=1;i<=m;i++)
    {
        int flag=0,temp=pe[i].p,cnt=0;
        while(temp>1)
        {
            if(mp[temp])
            {
                flag=1;
                mp[temp]--;
                break;
            }
            if(temp&1)temp=temp/2+1;
            else temp=temp/2;
            cnt++;
        }
        if(temp==1&&flag==0)
        {
            if(mp[1])flag=1,mp[1]--;
        }
        if(flag)
        {
            c++;u+=cnt;
            a[pe[i].id]=cnt;
            ha[num].p=temp;
            ha[num++].id=pe[i].id;
        }
    }
    printf("%d %d\n",c,u);
    for(int i=1;i<=m;i++)printf("%d ",a[i]);printf("\n");
    sort(ha,ha+num,cmp);
    int pos=0;
    for(int i=1;i<=n;i++)
    {
        if(ha[pos].p==po[i].p)
        {
            b[po[i].id]=ha[pos].id;
            pos++;
        }
        else b[po[i].id]=0;
    }
    for(int i=1;i<=n;i++)printf("%d ",b[i]);
    return 0;
}
/*
题意:只有能量值相等才能相连,现在可以让一个插座的能量值/2向上取整;问最多可以连接多少个电脑，在这个状态下需要向下多少次
以及具体的连接情况可操作次数的情况;
思路：贪心就好,从小到大开始/2然后搞搞就好了;
*/

　　

F：

题意：将无向边定向,使得最小的f(x)最大,f(x)表示x节点能到达的节点数目;

思路：可以发现当无向图中有f(x)最小的那个节点是最大的的那个边双连通分量里面的点,贪心可以知道先找边双连通分量，然后所有的点都指向这个边双连通分量;

AC代码：

#include <bits/stdc++.h>
using namespace std;
const int maxn=4e5+5;
int n,m,u[maxn],v[maxn],Vis[maxn],vis[maxn];
int low[maxn],pre[maxn],cnt,isbri[maxn],bcc_cnt=0,bcc[maxn];
struct Edge
{
    int to,id;
};
vector<Edge>ve[maxn];
int dfs(int cur,int fa)
{
    int tep=pre[cur]=++cnt;
    int len=ve[cur].size();
    for(int i=0;i<len;i++)
    {
        int x=ve[cur][i].to,y=ve[cur][i].id;
        if(x==fa)continue;
        if(!pre[x])
        {
            int temp=dfs(x,cur);
            tep=min(tep,temp);
            if(temp>pre[cur])isbri[y]=1;
        }
        else tep=min(tep,pre[x]);
    }
    low[cur]=tep;
    return tep;
}
void dfs1(int cur,int fa)
{
    vis[cur]=1;
    int len=ve[cur].size();
    for(int i=0;i<len;i++)
    {
        int x=ve[cur][i].to,y=ve[cur][i].id;
        if(x==fa||isbri[y]||vis[x])continue;
        bcc[bcc_cnt]++;
        Vis[y]=1;
        dfs1(x,cur);
    }
}
pair<int,int> pa[maxn];
int flag[maxn];
void dfs2(int cur,int fa)
{
    if(vis[cur])return ;
    vis[cur]=1;
    int len=ve[cur].size();
    for(int i=0;i<len;i++)
    {
        int x=ve[cur][i].to,y=ve[cur][i].id;
        if(x==fa)continue;
        if(!flag[y])
        {
            if(isbri[y])pa[y].first=x,pa[y].second=cur;
            else pa[y].first=cur,pa[y].second=x;
            flag[y]=1;
        }
        dfs2(x,cur);
    }
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d",&u[i],&v[i]);
        ve[u[i]].push_back((Edge){v[i],i});
        ve[v[i]].push_back((Edge){u[i],i});
    }
    if(m==n-1)
    {
        printf("1\n");
        for(int i=1;i<=m;i++)printf("%d %d\n",u[i],v[i]);
        return 0;
    }
    cnt=0;
    dfs(1,0);
    int ans=0,p;
    for(int i=1;i<=m;i++)
    {
        if(isbri[i]||Vis[i])continue;
        bcc[++bcc_cnt]=1;
        dfs1(u[i],0);
        if(bcc[bcc_cnt]>ans)
        {
            ans=bcc[bcc_cnt];
            p=u[i];
        }
    }
    memset(vis,0,sizeof(vis));
    dfs2(p,0);
    printf("%d\n",ans);
    for(int i=1;i<=m;i++)printf("%d %d\n",pa[i].first,pa[i].second);
    return 0;
}