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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  long long* arr = new long long[n];
  for (long long i = 0; i < n; ++i) {
    cin >> arr[i];
  }
  if (n == 1) {
    cout << arr[0] << endl;
    return 0;
  }
  if (n == 2) {
    cout << max(arr[0], arr[1]) << endl;
    return 0;
  }
  long long* large = new long long[n]();
  long long* secla = new long long[n]();
  large[0] = arr[0];
  large[1] = max(arr[0], arr[1]);
  secla[1] = min(arr[0], arr[1]);
  for (long long i = 2; i < n; ++i) {
    large[i] = max(large[i - 1], arr[i]);
    if (arr[i] == large[i]) {
      secla[i] = large[i - 1];
    } else {
      secla[i] = max(secla[i - 1], arr[i]);
    }
  }
  long long ans = INT_MAX;
  map<long long, long long> m;
  long long big = 0;
  for (long long i = 1; i < n; ++i) {
    if (large[i - 1] < arr[i]) {
      continue;
    } else if (secla[i - 1] < arr[i]) {
      ++m[large[i - 1]];
      big = max(big, m[large[i - 1]] - 1);
    } else {
      ans = min(ans, arr[i]);
    }
  }
  if (big == 0) {
    if (ans == INT_MAX) {
      cout << arr[0] << endl;
      return 0;
    }
    cout << ans << endl;
    return 0;
  }
  map<long long, long long>::iterator it = m.begin();
  while (it != m.end()) {
    if (it->second == big + 1) {
      cout << it->first << endl;
      return 0;
    }
    ++it;
  }
}
