B. Table Tennis

n people are standing in a line to play table tennis. At first, the first two players in the line play a game. Then the loser goes to the end of the line, and the winner plays with the next person from the line, and so on. They play until someone wins k games in a row. This player becomes the winner.

For each of the participants, you know the power to play table tennis, and for all players these values are different. In a game the player with greater power always wins. Determine who will be the winner.

Input

The first line contains two integers: n and k (2 ≤ n ≤ 500, 2 ≤ k ≤ 1012) — the number of people and the number of wins.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ n) — powers of the player. It's guaranteed that this line contains a valid permutation, i.e. all ai are distinct.

Output

Output a single integer — power of the winner.

    Examples 
  
      input 
    
2 2
1 2

      output 
    
2 

      input 
    
4 2
3 1 2 4

      output 
    
3 

      input 
    
6 2
6 5 3 1 2 4

      output 
    
6 

      input 
    
2 10000000000
2 1

      output 
    
2

Note

Games in the second sample:

3 plays with 1. 3 wins. 1 goes to the end of the line.

3 plays with 2. 3 wins. He wins twice in a row. He becomes the winner.

题意：

有1-n组成的一个排列，每个数代表一个人的power，两个人比赛时power大的人会获胜

比赛从左到右比，输了的人移到末尾，赢的人和下一个人比，当有一个人赢了k盘时输出该人的power

#include <iostream>

using namespace std;

int main()
{
    long long n,k;
    int fg=0,a[505];
    cin>>n>>k;
    for(int i=0;i<n;i++)
    cin>>a[i];
    if(k>=n)   //只有power为n的人才能赢>=n盘 ，所以k>=n时输出n
    cout<<n<<endl;
    else
    {
       for(int i=0;i<n;i++)
       {
          int s=0;
          if(i!=0&&a[i]>a[i-1])
          s=1;
          for(int j=1;j<=k-s;j++)
          {
             int pos=(i+j)%n;  //亮点，细细体会
             if(a[i]<a[pos])
             break;
             if(j==k-s)
             {
                fg=1;
                cout<<a[i]<<endl;
                break;
             }
          }
          if(fg)
          break;
       }
    }
    return 0;
}

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