Codeforces Round #408 (Div. 2) A B C 模拟 模拟 set

A. Buying A House

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Zane the wizard had never loved anyone before, until he fell in love with a girl, whose name remains unknown to us.

The girl lives in house m of a village. There are n houses in that village, lining in a straight line from left to right: house 1, house 2, ..., house n. The village is also well-structured: house i and house i + 1 (1 ≤ i < n) are exactly 10 meters away. In this village, some houses are occupied, and some are not. Indeed, unoccupied houses can be purchased.

You will be given n integers a₁, a₂, ..., a_n that denote the availability and the prices of the houses. If house i is occupied, and therefore cannot be bought, then a_i equals 0. Otherwise, house i can be bought, and a_i represents the money required to buy it, in dollars.

As Zane has only k dollars to spare, it becomes a challenge for him to choose the house to purchase, so that he could live as near as possible to his crush. Help Zane determine the minimum distance from his crush's house to some house he can afford, to help him succeed in his love.

Input

The first line contains three integers n, m, and k (2 ≤ n ≤ 100, 1 ≤ m ≤ n, 1 ≤ k ≤ 100) — the number of houses in the village, the house where the girl lives, and the amount of money Zane has (in dollars), respectively.

The second line contains n integers a₁, a₂, ..., a_n (0 ≤ a_i ≤ 100) — denoting the availability and the prices of the houses.

It is guaranteed that a_m = 0 and that it is possible to purchase some house with no more than k dollars.

Output

Print one integer — the minimum distance, in meters, from the house where the girl Zane likes lives to the house Zane can buy.

Examples

Input

5 1 20
0 27 32 21 19

Output

40

Input

7 3 50
62 0 0 0 99 33 22

Output

30

Input

10 5 100
1 0 1 0 0 0 0 0 1 1

Output

20

Note

In the first sample, with k = 20 dollars, Zane can buy only house 5. The distance from house m = 1 to house 5 is 10 + 10 + 10 + 10 = 40 meters.

In the second sample, Zane can buy houses 6 and 7. It is better to buy house 6 than house 7, since house m = 3 and house 6 are only 30 meters away, while house m = 3 and house 7 are 40 meters away.

题意：给你n m k  依次给你n个房子的价格 现在找离第m个房子最近的 价格不超过k 的距离 两个相邻的房子之间的距离为10

题解：水

 #include<bits/stdc++.h>
 #define ll __int64
 using namespace std;
 int n,m,k;
 int a[];
 int main()
 {
     scanf("%d %d %d",&n,&m,&k);
     for(int i=;i<=n;i++)
         scanf("%d",&a[i]);
     int ans=;
     for(int i=m+;i<=n;i++)
     {
         if(a[i]<=k&&a[i]!=)
         {
             ans=(i-m)*;
             break;
         }
     }
     for(int i=;i<m;i++)
     {
          if(a[i]<=k&&a[i]!=)
         {
             ans=min(ans,(m-i)*);
         }
     }
     printf("%d\n",ans);
     return ;
 }

B. Find The Bone

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Zane the wizard is going to perform a magic show shuffling the cups.

There are n cups, numbered from 1 to n, placed along the x-axis on a table that has m holes on it. More precisely, cup i is on the table at the position x = i.

The problematic bone is initially at the position x = 1. Zane will confuse the audience by swapping the cups k times, the i-th time of which involves the cups at the positions x = u_i and x = v_i. If the bone happens to be at the position where there is a hole at any time, it will fall into the hole onto the ground and will not be affected by future swapping operations.

Do not forget that Zane is a wizard. When he swaps the cups, he does not move them ordinarily. Instead, he teleports the cups (along with the bone, if it is inside) to the intended positions. Therefore, for example, when he swaps the cup at x = 4 and the one at x = 6, they will not be at the position x = 5 at any moment during the operation.

Zane’s puppy, Inzane, is in trouble. Zane is away on his vacation, and Inzane cannot find his beloved bone, as it would be too exhausting to try opening all the cups. Inzane knows that the Codeforces community has successfully helped Zane, so he wants to see if it could help him solve his problem too. Help Inzane determine the final position of the bone.

Input

The first line contains three integers n, m, and k (2 ≤ n ≤ 10⁶, 1 ≤ m ≤ n, 1 ≤ k ≤ 3·10⁵) — the number of cups, the number of holes on the table, and the number of swapping operations, respectively.

The second line contains m distinct integers h₁, h₂, ..., h_m (1 ≤ h_i ≤ n) — the positions along the x-axis where there is a hole on the table.

Each of the next k lines contains two integers u_i and v_i (1 ≤ u_i, v_i ≤ n, u_i ≠ v_i) — the positions of the cups to be swapped.

Output

Print one integer — the final position along the x-axis of the bone.

Examples

Input

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

Output

1

Input

5 1 2
2
1 2
2 4

Output

2

Note

In the first sample, after the operations, the bone becomes at x = 2, x = 5, x = 7, and x = 1, respectively.

In the second sample, after the first operation, the bone becomes at x = 2, and falls into the hole onto the ground.

题意： 桌子上倒着摆着n个杯子  初始骨头放在第一个杯子下 有m个杯子下有洞  骨头会掉入洞中 k次操作 每次交换两个杯子的位置(连同可能有的骨头) 问你经过k次操作之后 骨头在哪个杯子下面？

题解：模拟 注意第一个杯子下有洞的情况。

 #include<bits/stdc++.h>
 #define ll __int64
 using namespace std;
 int n,m,k;
 int a[];
 int x,y;
 map<int,int>mp;
 map<int,int>mpp;
 int main()
 {
     scanf("%d %d %d",&n,&m,&k);
     for(int i=; i<=m; i++)
     {
         scanf("%d",&a[i]);
         mp[a[i]]=;
     }
     int flag=;
     int now=-;
     mpp[]=;
     int ff=;
     for(int i=; i<=k; i++)
     {
         scanf("%d %d",&x,&y);
         if(mpp[x]==&&mpp[y]==)
             continue;
         else
         {
             if(mpp[x]==)
             {
                 if(mp[y]==&&flag==)
                 {
                     now=y;
                     flag=;
                 }
                 else
                 {
                     mpp[x]=;
                     mpp[y]=;
                     ff=y;
                 }
             }
             else
             {
                 if(mp[x]==&&flag==)
                 {
                     now=x;
                     flag=;
                 }
                 else
                 {
                     mpp[y]=;
                     mpp[x]=;
                     ff=x;
                 }
             }
         }
     }
         if(mp[]==)
         {
             printf("1\n");
             return ;
         }

         if(now==-)
             printf("%d\n",ff);
         else
             printf("%d\n",now);
         return ;
     }
 /*
 7 3 4
 1 4 6
 1 2
 2 5
 5 7
 7 1
 */

C. Bank Hacking

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Although Inzane successfully found his beloved bone, Zane, his owner, has yet to return. To search for Zane, he would need a lot of money, of which he sadly has none. To deal with the problem, he has decided to hack the banks.

There are n banks, numbered from 1 to n. There are also n - 1 wires connecting the banks. All banks are initially online. Each bank also has its initial strength: bank i has initial strength a_i.

Let us define some keywords before we proceed. Bank i and bank j are neighboring if and only if there exists a wire directly connecting them. Bank i and bank j are semi-neighboring if and only if there exists an online bank k such that bank i and bank k are neighboring and bank k and bank j are neighboring.

When a bank is hacked, it becomes offline (and no longer online), and other banks that are neighboring or semi-neighboring to it have their strengths increased by 1.

To start his plan, Inzane will choose a bank to hack first. Indeed, the strength of such bank must not exceed the strength of his computer. After this, he will repeatedly choose some bank to hack next until all the banks are hacked, but he can continue to hack bank x if and only if all these conditions are met:

1. Bank x is online. That is, bank x is not hacked yet.
2. Bank x is neighboring to some offline bank.
3. The strength of bank x is less than or equal to the strength of Inzane's computer.

Determine the minimum strength of the computer Inzane needs to hack all the banks.

Input

The first line contains one integer n (1 ≤ n ≤ 3·10⁵) — the total number of banks.

The second line contains n integers a₁, a₂, ..., a_n ( - 10⁹ ≤ a_i ≤ 10⁹) — the strengths of the banks.

Each of the next n - 1 lines contains two integers u_i and v_i (1 ≤ u_i, v_i ≤ n, u_i ≠ v_i) — meaning that there is a wire directly connecting banks u_i and v_i.

It is guaranteed that the wires connect the banks in such a way that Inzane can somehow hack all the banks using a computer with appropriate strength.

Output

Print one integer — the minimum strength of the computer Inzane needs to accomplish the goal.

Examples

Input

5
1 2 3 4 5
1 2
2 3
3 4
4 5

Output

5

Input

7
38 -29 87 93 39 28 -55
1 2
2 5
3 2
2 4
1 7
7 6

Output

93

Input

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

Output

8

Note

In the first sample, Inzane can hack all banks using a computer with strength 5. Here is how:

• Initially, strengths of the banks are [1, 2, 3, 4, 5].
• He hacks bank 5, then strengths of the banks become [1, 2, 4, 5,  - ].
• He hacks bank 4, then strengths of the banks become [1, 3, 5,  - ,  - ].
• He hacks bank 3, then strengths of the banks become [2, 4,  - ,  - ,  - ].
• He hacks bank 2, then strengths of the banks become [3,  - ,  - ,  - ,  - ].
• He completes his goal by hacking bank 1.

In the second sample, Inzane can hack banks 4, 2, 3, 1, 5, 7, and 6, in this order. This way, he can hack all banks using a computer with strength 93.

题意：给你一颗树 有点权  初始选取一个点i进行hack操作  与i相连的点 点权增加1  之后选取与i相连的所有点j进行hack操作 同样与j相连的点 点权增加1(点被hack掉之后权值不再变化) 直到将n个点全部hack掉 进行hack操作 需要一个力量值大于等于当前点的点权 问最小的力量值hack掉所有的点。

题解：枚举初始hack点i   ans=min(ans,max(i的点权，max(与i相连的点的权值最大值+1，其余点的权值最大值+2)))；  set 处理；

 #include<bits/stdc++.h>
 #define ll __int64
 using namespace std;
 int n;
 ll a[];
 vector<ll> mp[];
 multiset<ll> s;
 multiset<ll>::iterator it;
 int main()
 {
     scanf("%d",&n);
     for(int i=; i<=n; i++)
     {
         scanf("%I64d",&a[i]);
         s.insert(a[i]);
     }
     ll l,r;
     for(int i=; i<n; i++)
     {
         scanf("%I64d %I64d",&l,&r);
         mp[l].push_back(r);
         mp[r].push_back(l);
     }
     ll ans=;
     ll exm1=-1e9-,exm2=-1e9-;
     for(int i=; i<=n; i++)
     {
         exm1=-1e9-;
         exm2=-1e9-;
         it=s.find(a[i]);
         s.erase(it);
         for(int j=; j<mp[i].size(); j++)
         {
             exm1=max(exm1,a[mp[i][j]]);
             it=s.find(a[mp[i][j]]);
             s.erase(it);
         }
         if(s.empty())
         {
             ans=min(ans,max(a[i],exm1+));
             for(int j=; j<mp[i].size(); j++)
                 s.insert(a[mp[i][j]]);
             s.insert(a[i]);
             continue;
         }
         it=s.end();
         it--;
         exm2=*it;
         ans=min(ans,max(max(a[i],exm1+),exm2+));
         for(int j=; j<mp[i].size(); j++)
             s.insert(a[mp[i][j]]);
         s.insert(a[i]);
     }
     printf("%I64d\n",ans);
     return ;
 }