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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, a[100001] = {}, cnt[100001] = {};
  cin >> n;
  int cur = 0, mx = 0;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    if (a[i] > a[cur]) {
      mx = a[cur];
      cur = i;
      continue;
    }
    if (a[i] > mx) {
      cnt[a[cur]]++;
      mx = a[i];
    }
  }
  int ans = 0;
  mx = 1;
  for (int i = 1; i <= n; ++i) {
    if (cnt[i] > mx) {
      mx = cnt[i];
      ans = i;
    }
  }
  mx = 0;
  if (ans == 0) {
    ans = n + 1;
    for (int i = 1; i <= n; ++i) {
      if (a[i] <= mx) ans = min(ans, a[i]);
      mx = max(mx, a[i]);
    }
  }
  cout << (ans == n + 1 ? 1 : ans) << endl;
  return 0;
}
