Codeforces 900C-Remove Extra One

Remove Extra One

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a permutation p of length n. Remove one element from permutation to make the number of records the maximum possible.

We remind that in a sequence of numbers a1, a2, ..., ak the element ai is a record if for every integer j (1 ≤ j < i) the following holds: aj < ai.

Input

The first line contains the only integer n (1 ≤ n ≤ 105) — the length of the permutation.

The second line contains n integers p1, p2, ..., pn (1 ≤ pi ≤ n) — the permutation. All the integers are distinct.

Output

Print the only integer — the element that should be removed to make the number of records the maximum possible. If there are multiple such elements, print the smallest one.

Examples

input

11

output

1

input

55 1 2 3 4

output

5

Note

In the first example the only element can be removed.

题意：给一个N个数的排列，要求移走一个数，使得record数最多。record数就是这个数比排在它前面的数都大

解题思路：如果a[i]如果是a[1..i]中最大的数字那么record会减少1。对于任意一个a[j],且i<j且a[i]是a[1..j]中最大的数字,且a[j]是a[1..j]中第二大的数字，删掉a[i]后，a[j]都会成为一个record,即record数递增

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>#include <ctime>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;int n, a[100009], cnt[100009];int main(){while (~scanf("%d", &n)){for (int i = 1; i <= n; i++) scanf("%d", &a[i]);memset(cnt, 0, sizeof cnt);int mx1 = 0, mx2 = 0;for (int i = 1; i <= n; i++){if (a[i] > mx1){mx2 = mx1;mx1 = a[i];cnt[a[i]]--;}else if (a[i] > mx2){cnt[mx1]++;mx2 = a[i];}}int ans, mx = -10;for (int i = 1; i <= n; i++)if (cnt[i] > mx) mx = cnt[i], ans = i;printf("%d\n", ans);}return 0;}