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;
int main() {
  int n, maxx = 0, a1, a2, pref = 0, ans = 1, t = 0;
  scanf("%d", &n);
  vector<int> a(n), r(n, 0), x(n, 0);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n; i++) {
    if (a[i] > maxx) {
      r[i] = 1;
      maxx = a[i];
    }
    pref += r[i];
  }
  a1 = a[0];
  a2 = 0;
  for (int i = 1; i < n; i++)
    if (a[i] > a1) {
      a2 = a1;
      a1 = a[i];
      t = i;
    } else if (a[i] > a2) {
      a2 = a[i];
      if (r[i] == 0) x[t]++;
    }
  maxx = 0;
  for (int i = 0; i < n; i++) {
    int t = pref - r[i] + x[i];
    if (t > maxx) {
      maxx = t;
      ans = a[i];
    }
    if (t == maxx) ans = min(ans, a[i]);
  }
  printf("%d", ans);
  return 0;
}
