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 = 1e6 + 5;
int arr[MAXN], vids[MAXN];
int n;
int main() {
  while (~scanf("%d", &n)) {
    memset(arr, 0, sizeof(arr));
    memset(vids, 0, sizeof(vids));
    int max1, max2;
    max1 = max2 = 0;
    for (int i = 0; i < n; i++) {
      scanf("%d", &arr[i]);
      if (arr[i] > max1) {
        vids[arr[i]]--;
        max2 = max1;
        max1 = arr[i];
      } else if (arr[i] >= max2 && arr[i] <= max1) {
        vids[max1]++;
        max2 = arr[i];
      }
    }
    int re = 0;
    int maxs = -0x7fffffff;
    for (int i = 0; i < n; i++) {
      if (vids[arr[i]] > maxs) {
        re = arr[i];
        maxs = vids[arr[i]];
      } else if (vids[arr[i]] == maxs) {
        if (re > arr[i]) re = arr[i];
      }
    }
    printf("%d\n", re);
  }
  return 0;
}
