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
Copy
2 2
1 2
output
2
input
Copy
4 2
3 1 2 4
output
3
input
Copy
6 2
6 5 3 1 2 4
output
6
input
Copy
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.
简单题，注意k很大，当k>n时，输出最大的那个就行
#include <bits/stdc++.h>
using namespace std;
#define maxn 100000
typedef long long ll;
#define inf 2147483647
#define ri register int
vector<ll> vec;
ll p[maxn];
ll cnt[maxn];
ll n, k;
ll ma = ; int main() {
// freopen("test.txt", "r", stdin);
cin >> n >> k;
for (int i = ; i <= n; i++) {
cin >> p[i];
ma = max(ma, p[i]);
vec.push_back(i);
}
if (k > n) {
cout << ma;
return ;
}
while () {
// for(ll i:vec)cout<<i<<" ";
// cout<<endl;
int t1 = vec.front();
vec.erase(vec.begin());
int t2 = vec.front();
vec.erase(vec.begin());
if (p[t1] > p[t2]) {
cnt[t1]++;
if (cnt[t1] == k) {
cout << p[t1];
return ;
}
vec.push_back(t2);
vec.insert(vec.begin(), t1);
} else {
cnt[t2]++;
if (cnt[t2] == k) {
cout << p[t2];
return ;
}
vec.push_back(t1);
vec.insert(vec.begin(), t2);
}
} return ;
}