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;
using ll = long long;
using ull = unsigned long long;
int arr[100005];
bool record[100005];
int rem[100005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  set<int> s;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    auto it = s.upper_bound(arr[i]);
    if (it == s.end())
      record[arr[i]] = true;
    else if (next(it) == s.end())
      rem[*it]++;
    s.insert(arr[i]);
  }
  int cnt = 0;
  for (int i = 1; i <= n; i++)
    if (record[i]) cnt++;
  int choose = -1;
  int meas = -2;
  for (int i = 0; i < n; i++) {
    int res = cnt + rem[arr[i]];
    if (record[arr[i]]) res--;
    if (res > meas) {
      meas = res;
      choose = arr[i];
    } else if (res == meas) {
      choose = min(choose, arr[i]);
    }
  }
  cout << choose << "\n";
  cerr << "Execution time: " << 1.0 * clock() / CLOCKS_PER_SEC << "s.\n";
  return 0;
}
