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 MAX_N = 100005;
int cnt[MAX_N];
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> vec(n);
  for (int i = 0; i < (int)(n); ++i) {
    cin >> vec[i];
  }
  int m1 = -1, m2 = -2;
  int id1 = -1, id2 = -1;
  for (int i = 0; i < (int)(n); ++i) {
    if (vec[i] > m1) {
      cnt[i] -= 1;
      m2 = m1, id2 = id1;
      m1 = vec[i], id1 = i;
    } else if (vec[i] > m2) {
      cnt[id1] += 1;
      m2 = vec[i], id2 = i;
    }
  }
  vector<pair<int, int> > res;
  for (int i = 0; i < (int)(n); ++i) {
    res.push_back(pair<int, int>(cnt[i], -vec[i]));
  }
  sort((res).begin(), (res).end(), greater<pair<int, int> >());
  cout << -res[0].second << "\n";
  return 0;
}
