CodeForces - 879B Table Tennis （不要只顾这个数的一边，这个数的前面和后面都的顾）

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, ..., a_n (1 ≤ a_i ≤ n) — powers of the player. It's guaranteed that this line contains a valid permutation, i.e. all a_i are distinct.

Output

Output a single integer — power of the winner.

Example

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： 打乒乓球，输入两个数 n，h，共有n 个人，h为要赢的场次数，下面一行为n个人的能量；先让前两个人开始打比赛，赢的人依次和下面的人打比赛，输的人在最后面排队，两个人输赢有两个人的能量定，能量高的人赢，输出第一个赢的h场的人；

思路：这是一道模拟题，数据范围可定义为 long long 首先要思考若轮完一轮了，还没有人赢h场，这时候一定是能量最高的那个人赢；还有在中间的人一定要和他的前一个人比（这一不要忘了）也和他后面的人比赛；记录每一个人第一轮能赢几场就行了，这道题主要就是不要只顾一边，前 后 都要顾，一定不要忘了和他前一个人相比；

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define Max 550

long long a[Max],hou[Max];
int main()
{
	long long i,j,n,h;
	while(~scanf("%lld%lld",&n,&h))
	{
		long long Ma = -1;
		for(i=0;i<n;i++)
		{
			scanf("%lld",&a[i]);
			Ma = max(Ma,a[i]);
		}	
		for(i=0;i<n;i++)
		{
			long long num ;
			if(i>0&&a[i]>a[i-1])  // 一定不要忘了和他前一个人相比； 
				num = 1;
			else num = 0;
			for(j=i+1;j<n;j++)
			{
				if(a[i]>a[j])
					num++;
				else break;
			}
			if(num>=h)
			{
				printf("%lld\n",a[i]);
				break;
			}
		}
		if(i>=n)
			printf("%lld\n",Ma);
	}
	return 0;
}