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 mpow(int base, int exp);
void ipgraph(int n, int m);
void dfs(int u, int par);
const int N = 3e5, M = N;
void solve() {
  int n;
  cin >> n;
  vector<int> arr(n);
  for (int i = 0; i < (n); i++) cin >> arr[i];
  vector<int> count(n, 0);
  set<int> m;
  m.insert(arr[0]);
  vector<int> isrecord(n, false);
  isrecord[arr[0] - 1] = true;
  int records = 1;
  for (int i = (1); i <= (n - 1); i++) {
    m.insert(arr[i]);
    if (*m.rbegin() == arr[i]) {
      records++;
      isrecord[arr[i] - 1] = true;
      continue;
    } else if (*(++m.rbegin()) == arr[i])
      count[*m.rbegin() - 1]++;
  }
  int mx = INT_MIN, ans;
  for (int i = 0; i < (n); i++) {
    int curr_records = records + count[i] - isrecord[i];
    if (mx < curr_records) {
      mx = curr_records;
      ans = i;
    } else if (mx == curr_records)
      ans = min(ans, i);
  }
  cout << ans + 1 << endl;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  srand(chrono::high_resolution_clock::now().time_since_epoch().count());
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
