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;
const long long int oo = 1e18;
const long long int MOD = 1e9 + 7;
const long long int maxN = 2e5 + 5;
int a[maxN], n, res[maxN];
void solve() {
  cin >> n;
  int max1 = 0, max2 = 0, pos = 0;
  for (long long int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] > max1) {
      max2 = max1;
      max1 = a[i];
      pos = i;
      res[i] = -1;
    } else if (a[i] > max2) {
      res[pos]++;
      max2 = a[i];
    }
  }
  int ans = 1;
  for (long long int i = 2; i <= n; i++)
    if (res[i] > res[ans])
      ans = i;
    else if (res[i] == res[ans] && a[i] < a[ans])
      ans = i;
  cout << a[ans];
}
int main() {
  {
    ios::sync_with_stdio(false);
    cin.tie(0);
  };
  int t = 1;
  while (t--) solve();
  return 0;
}
