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;
long long a[100050];
long long effect[100050];
int main() {
  long long i, j, k, n, m, max1, max2, maxid, ans;
  while (scanf("%lld", &n) != EOF) {
    memset(effect, 0, sizeof(effect));
    for (i = 1; i <= n; i++) scanf("%lld", &a[i]);
    bool flag = true;
    for (i = 1; i <= n; i++)
      if (a[i] != i) {
        flag = false;
        break;
      }
    if (n == 1 || n == 2 || flag) {
      printf("1\n");
      continue;
    }
    if (a[1] > a[2]) {
      max1 = a[1];
      max2 = a[2];
      maxid = 1;
      effect[1] = 1;
    } else {
      max1 = a[2];
      max2 = a[1];
      maxid = 2;
    }
    for (i = 3; i <= n; i++) {
      if (a[i] > max1) {
        maxid = i;
        max2 = max1;
        max1 = a[i];
      } else {
        if (a[i] > max2) {
          effect[maxid]++;
          max2 = a[i];
        }
      }
    }
    for (i = 2; i <= n; i++)
      if (effect[i] > 0) {
        effect[i]--;
      }
    effect[1]--;
    ans = 1;
    for (i = 2; i <= n; i++)
      if (effect[i] > effect[ans])
        ans = i;
      else if (effect[i] == effect[ans] && a[i] < a[ans])
        ans = i;
    printf("%lld\n", a[ans]);
  }
}
