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 int N = 1e5;
map<int, int> dic;
int n, a[N + 10], ma[N + 10], ma2[N + 10];
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  ma[1] = a[1];
  ma2[1] = -1;
  for (int i = 2; i <= n; i++) {
    if (a[i] > ma[i - 1]) {
      ma2[i] = ma[i - 1];
      ma[i] = a[i];
    } else {
      ma[i] = ma[i - 1];
      if (ma2[i - 1] < a[i]) {
        ma2[i] = a[i];
      } else
        ma2[i] = ma2[i - 1];
    }
    if (a[i] != ma[i] && a[i] == ma2[i]) {
      dic[ma[i]]++;
    }
  }
  int tot = -1e6, idx = -1;
  for (int i = 1; i <= n; i++) {
    int temp = dic[a[i]];
    if (a[i] == ma[i]) temp--;
    if (temp > tot) {
      tot = temp;
      idx = a[i];
    } else if (temp == tot) {
      idx = min(idx, a[i]);
    }
  }
  cout << idx << endl;
  return 0;
}
