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 v[200100];
int n;
set<int> s;
int a[200100];
int b[200100];
int inv[200100];
int ans = -0x3F3F3F3F;
int cnt;
int id;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &v[i]);
    inv[v[i]] = i;
    if (s.size() == 0) {
      a[i] = 1;
      cnt++;
    }
    if (s.size() == 1) {
      if (*s.begin() < v[i]) {
        a[i] = 1;
        cnt++;
      } else {
        b[inv[*s.begin()]]++;
      }
    }
    if (s.size() >= 2) {
      if (*s.rbegin() < v[i]) {
        a[i] = 1;
        cnt++;
      } else {
        int aux = *s.rbegin();
        s.erase(*s.rbegin());
        if (*s.rbegin() < v[i]) {
          b[inv[aux]]++;
        }
        s.insert(aux);
      }
    }
    s.insert(v[i]);
  }
  for (int i = 0; i < n; ++i) {
    int aux = cnt - a[i] + b[i];
    if (aux > ans) {
      ans = aux;
      id = v[i];
    } else if (aux == ans) {
      id = min(id, v[i]);
    }
  }
  printf("%d\n", id);
  return 0;
}
