Codeforces Round #443 (Div.2)

B. Table Tennis

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 21 2

output

2 

input

4 23 1 2 4

output

3 

input

6 26 5 3 1 2 4

output

6 

input

2 100000000002 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

思路：

只有power为n的人才能赢>=n盘 ，所以k>=n时输出n

k<n时从左到右进行遍历，遇到可以赢k盘的人就结束。（如果这个人的power比前一个人的power要来得大，接下来只要再赢k-1盘）

#include<iostream>#include<string.h>#include<stdio.h>using namespace std;#define LL long longconst LL maxn = 555;LL n,k,a[maxn];int main(){    scanf("%lld%lld",&n,&k);    for(int i=0;i<n;i++) {        scanf("%lld",&a[i]);    }    if(k>=n){        printf("%lld\n",n);        return 0;    }    for(int i=0;i<n;i++){        if(a[i]<k) continue;        int add = 0;        if(i!=0&&a[i-1]<a[i]) add = 1;        for(int j=1;j<=k-add;j++){            int pos = (i+j)%n;            if(a[i]<a[pos]) break;            if(j==k-add){                printf("%lld\n",a[i]);                return 0;            }        }    }    return 0;}