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 a[100100], b[100100], ans, pp;
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  if (n == 1) {
    cout << a[1];
    return 0;
  }
  int pos = 1;
  int bef, count, ma = 0;
  while (pos <= n)
    if (a[pos] < a[pos + 1])
      b[pos++] = 1;
    else {
      count = 0;
      bef = (a[pos - 1] < ma ? ma : a[pos - 1]);
      ma = a[pos];
      b[pos++] = 1;
      while (a[pos] <= ma) {
        if (pos > n) break;
        if (a[pos] > bef) count++, bef = a[pos];
        pos++;
      }
      if (count > ans) ans = count, pp = ma;
    }
  ma = n + 1;
  for (int i = 1; i <= n; i++)
    if (!b[i]) ma = (ma < a[i] ? ma : a[i]);
  if (ans == 0)
    if (ma == n + 1)
      pp = a[1];
    else
      pp = ma;
  if (ans == 1) pp = (ma < pp ? ma : pp);
  cout << pp;
  return 0;
}
