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::sync_with_stdio(false);
  int n;
  cin >> n;
  int arr[n];
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  int cnt[n + 1], isr[n + 1];
  for (int i = 0; i <= n; i++) {
    cnt[i] = 0;
    isr[i] = -1;
  }
  set<int> s;
  s.insert(arr[0]);
  isr[arr[0]] = 1;
  for (int i = 1; i < n; i++) {
    set<int>::iterator itr;
    s.insert(arr[i]);
    itr = s.find(arr[i]);
    itr++;
    if (itr != s.end()) {
      itr++;
      if (itr == s.end()) {
        itr--;
        cnt[arr[i]] = *itr;
      }
    } else {
      isr[arr[i]] = 1;
    }
  }
  int cnt2[n + 1];
  for (int i = 0; i <= n; i++) cnt2[i] = 0;
  for (int i = 1; i <= n; i++) {
    if (cnt[i] != 0) {
      cnt2[cnt[i]]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (isr[i] == 1 and cnt2[i] != 0) {
      cnt2[i]--;
    }
  }
  int ans = 1, temp = 0;
  for (int i = 1; i <= n; i++) {
    if (isr[i] == -1) {
      ans = i;
      break;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (cnt2[i] > temp) {
      ans = i;
      temp = cnt2[i];
    }
  }
  cout << ans;
  return 0;
}
