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

1
1


Output

1


Input

5
5 1 2 3 4


Output

5

Note

In the first example the only element can be removed.
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100011;
int mx[maxn], mxn[maxn];
int a[maxn];
int cnt[maxn];
int main(int argc, char const *argv[]) {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    mx[i] = mx[i - 1];
    mxn[i] = mxn[i - 1];
    if (a[i] >= mx[i])
      mxn[i] = mx[i], mx[i] = a[i];
    else if (a[i] >= mxn[i])
      mxn[i] = a[i];
  }
  for (int i = 1; i <= n; i++)
    if (mx[i] == a[i]) cnt[a[i]]--;
  for (int i = 1; i <= n; i++)
    if (mxn[i] == a[i]) cnt[mx[i]]++;
  int ans = 1, now = -10;
  for (int i = 1; i <= n; i++)
    if (cnt[i] > now) now = cnt[i], ans = i;
  printf("%d\n", ans);
  return 0;
}
