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 po(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = (res * a) % 998244353;
    }
    a = (a * a) % 998244353;
    b = b / 2;
  }
  return res % 998244353;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  {
    int n, a = 0, m = 0, next = 0, ans = 1, pre[100100];
    cin >> n;
    for (int i = 0; i < n; i++) {
      cin >> a;
      if (a > m)
        pre[a] = 1, next = m, m = a;
      else if (a > next)
        pre[m]--, next = a;
    }
    for (int i = 1; i <= n; ++i)
      if (pre[i] < pre[ans]) ans = i;
    cout << ans;
  }
  return 0;
}
