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 long long int mod = 1e9 + 7;
void solve() {
  long long int n;
  cin >> n;
  long long int a[n];
  for (long long int i = 0; i < n; i++) cin >> a[i];
  map<long long int, long long int> m;
  for (long long int i = 0; i < n; i++) m[i + 1] = 0;
  long long int ma = -1, mb = -1;
  for (long long int i = 0; i < n; i++) {
    long long int x = a[i];
    if (x >= ma) {
      mb = ma;
      ma = x;
      m[x]--;
    } else if (x > mb) {
      mb = x;
      m[ma]++;
    }
  }
  {
    long long int ans = (*m.begin()).second, v = (*m.begin()).first;
    for (auto x : m) {
      if (x.second > ans) {
        v = x.first;
        ans = x.second;
      }
    }
    cout << v << "\n";
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  clock_t launch = clock();
  while (t--) {
    solve();
  }
  clog << ((long double)(clock() - launch) / CLOCKS_PER_SEC) << "\n";
}
