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 d[101000];
bool pr[101000];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int i, n, a, b, mx1, mx, mx2, mn = 1e9;
  cin >> n;
  if (n <= 2) {
    cout << 1;
    return 0;
  }
  cin >> a >> b;
  mx1 = max(a, b);
  mx2 = min(a, b);
  if (a < b)
    pr[a] = pr[b] = true;
  else
    pr[b] = true;
  if (b < a) d[a]++;
  for (i = 2; i < n; i++) {
    cin >> a;
    if (a < mx1 && a > mx2) d[mx1]++;
    if (a < mx2) mn = min(mn, a);
    if (a > mx1) {
      pr[a] = true;
      mx2 = mx1;
      mx1 = a;
    } else if (a > mx2)
      mx2 = a;
  }
  mx = 1;
  for (i = 2; i <= n; i++)
    if (d[i] > d[mx] && d[i] > 1 || d[i] == d[mx] && pr[mx] && !pr[i]) mx = i;
  cout << mx;
  return 0;
}
