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 MOD = 1e9 + 7;
const int N = 1e6 + 7;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> v(n), wrong(n, 0);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    v[i]--;
  }
  set<int> s, goods;
  for (int i = 0; i < n; i++) {
    s.insert(v[i]);
    auto it = s.upper_bound(v[i]);
    if (it == s.end()) {
      goods.insert(v[i]);
      continue;
    }
    int x = *it;
    it++;
    if (it == s.end()) wrong[x]++;
  }
  int mx = -1, res = 0;
  for (int i = 0; i < n; i++) {
    int tmp = goods.size() + wrong[i];
    if (goods.count(i)) tmp--;
    if (tmp > mx) {
      mx = tmp;
      res = i;
    }
  }
  cout << res + 1 << endl;
}
