Testing Round #12 A,B,C 讨论，贪心，树状数组优化dp

题目链接：http://codeforces.com/contest/597

A. Divisibility

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Find the number of k-divisible numbers on the segment [a, b]. In other words you need to find the number of such integer values x that a ≤ x ≤ b and x is divisible by k.

Input

The only line contains three space-separated integers k, a and b (1 ≤ k ≤ 1018; - 1018 ≤ a ≤ b ≤ 1018).

Output

Print the required number.

Examples

input

1 1 10

output

10

input

2 -4 4

output

5

题意：找出[a,b]区间内整除k的数的个数；

思路:小心点特判即可；

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=1e5+,M=4e6+,inf=;
const ll INF=1e18+,mod=1e9+;
///   数组大小
int main()
{
    ll k,a,b;
    scanf("%lld%lld%lld",&k,&a,&b);
    if(a<=&&b>=)
        printf("%lld\n",(b/k)-(a/k)+);
    else if(a>=&&b>=)
        printf("%lld\n",(b/k)-(a/k+(a%k?:))+);
    else
        printf("%lld\n",(abs(a)/k)-(abs(b)/k+(abs(b)%k?:))+);
    return ;
}

B. Restaurant

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A restaurant received n orders for the rental. Each rental order reserve the restaurant for a continuous period of time, the i-th order is characterized by two time values — the start time li and the finish time ri (li ≤ ri).

Restaurant management can accept and reject orders. What is the maximal number of orders the restaurant can accept?

No two accepted orders can intersect, i.e. they can't share even a moment of time. If one order ends in the moment other starts, they can't be accepted both.

Input

The first line contains integer number n (1 ≤ n ≤ 5·105) — number of orders. The following n lines contain integer values li and ri each (1 ≤ li ≤ ri ≤ 109).

Output

Print the maximal number of orders that can be accepted.

Examples

input

2
7 11
4 7

output

1

input

5
1 2
2 3
3 4
4 5
5 6

output

3

input

6
4 8
1 5
4 7
2 5
1 3
6 8

output

2

贪心：按r从小到大排序即可；

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=5e5+,M=4e6+,inf=;
const ll INF=1e18+,mod=1e9+;
struct is
{
    int l,r;
    bool operator <(const is &c)const
    {
        return r<c.r;
    }
}a[N];

///   数组大小

int main()
{
    int n;
    scanf("%d",&n);
    for(int i=;i<=n;i++)
        scanf("%d%d",&a[i].l,&a[i].r);
    sort(a+,a++n);
    int s=,ans=;
    for(int i=;i<=n;i++)
    {
        if(a[i].l>s)
        {
            ans++;
            s=a[i].r;
        }
    }
    printf("%d\n",ans);
    return ;
}

C. Subsequences

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

For the given sequence with n different elements find the number of increasing subsequences with k + 1 elements. It is guaranteed that the answer is not greater than 8·1018.

Input

First line contain two integer values n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ 10) — the length of sequence and the number of elements in increasing subsequences.

Next n lines contains one integer ai (1 ≤ ai ≤ n) each — elements of sequence. All values ai are different.

Output

Print one integer — the answer to the problem.

Examples

input

5 2
1
2
3
5
4

output

7

题意：给你n个数，最长上升子序列长度为k+1的个数；

思路：看下数据范围k<10很关键,dp[i][j]表示以i为结束长度为j时候的方案数

　　　现在你到i的时候你只需要再T[j]（树状数组)的a[i]的位置表示方案数；

　　　统计小于a[i]的方案，k==0时候特判；

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=1e5+,M=4e6+,inf=;
const ll INF=1e18+,mod=1e9+;
struct AYT
{
    ll tree[N];
    int lowbit(int x)
    {
        return x&-x;
    }
    void update(int x,ll c)
    {
        while(x<N)
        {
            tree[x]+=c;
            x+=lowbit(x);
        }
    }
    ll query(int x)
    {
        ll ans=;
        while(x)
        {
            ans+=tree[x];
            x-=lowbit(x);
        }
        return ans;
    }
};
AYT T[];

///   数组大小
int a[N];
int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    for(int i=;i<=n;i++)
        scanf("%d",&a[i]);
    if(k==)
        return *printf("%d\n",n);
    ll ans=;
    for(int i=;i<=n;i++)
    {
        ans+=T[k].query(a[i]-);
        for(int j=k;j>=;j--)
        {
            ll x=T[j-].query(a[i]-);
            T[j].update(a[i],x);
        }
        T[].update(a[i],);
    }
    printf("%lld\n",ans);
    return ;
}

A. Divisibility

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Find the number of k-divisible numbers on the segment [a, b]. In other words you need to find the number of such integer values x that a ≤ x ≤ b and x is divisible by k.

Input

The only line contains three space-separated integers k, a and b (1 ≤ k ≤ 1018; - 1018 ≤ a ≤ b ≤ 1018).

Output

Print the required number.

Examples

input

1 1 10

output

10

input

2 -4 4

output

5