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 PI = acos(-1.0);
const int MOD = 1e9 + 7;
const int MAX_N = 1e5 + 5;
int n;
int arr[MAX_N];
int if_removed_will_increase[MAX_N];
int max1[MAX_N];
int max2[MAX_N];
int min1 = MAX_N;
int is_record_initially[MAX_N];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  int initial_records = 0;
  for (int i = 0; i < n; ++i) {
    cin >> arr[i];
  }
  for (int i = 0; i < n; ++i) {
    min1 = min(min1, arr[i]);
  }
  set<int, greater<int> > s;
  s.insert(0);
  for (int i = 0; i < n; ++i) {
    s.insert(arr[i]);
    max1[i] = *(s.begin());
    max2[i] = *(++s.begin());
    if (arr[i] >= max1[i]) {
      is_record_initially[i] = true;
      ++initial_records;
    }
  }
  for (int i = 0; i < n; ++i) {
    if (arr[i] < max1[i] && arr[i] >= max2[i]) {
      ++if_removed_will_increase[max1[i]];
    }
  }
  int max_records = -1;
  int ans = max1[n - 1];
  for (int i = 0; i < n; ++i) {
    int if_removed = initial_records - is_record_initially[i] +
                     if_removed_will_increase[arr[i]];
    if (if_removed > max_records) {
      max_records = if_removed;
      ans = arr[i];
    } else if (if_removed == max_records) {
      ans = min(ans, arr[i]);
    }
  }
  cout << ans << '\n';
  return 0;
}
