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;
  cin >> n;
  vector<int> a(n), pmx(n, -1), smx(n, -1);
  cin >> a[0];
  pmx[0] = 0;
  int c = 1;
  for (int j = 1; j < n; j++) {
    cin >> a[j];
    pmx[j] = pmx[j - 1];
    smx[j] = (j == 1) + (j != 1) * smx[j - 1];
    if (a[j] > a[pmx[j]]) {
      smx[j] = pmx[j];
      pmx[j] = j;
      c++;
    } else if (a[j] > a[smx[j]]) {
      smx[j] = j;
    }
  }
  vector<int> rec(n, c);
  int mx = -1, i;
  for (int j = 0; j < n; j++) {
    rec[j] -= (pmx[j] == j);
  }
  for (int j = 1; j < n; j++) {
    if (smx[j] != j) continue;
    rec[pmx[j]]++;
  }
  for (int j = 0; j < n; j++) {
    if (mx < rec[j]) {
      mx = rec[j];
      i = j;
    } else if (mx == rec[j] && a[j] < i) {
      i = j;
    }
  }
  int mn = n;
  for (int j = 0; j < n; j++) {
    if (rec[i] == rec[j] && mn > a[j]) {
      mn = a[j];
    }
  }
  cout << mn;
}
