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 INF = 1e9 + 7;
const int MAXN = 1e6 + 7;
const double EPS = 1e-8;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  set<int> q;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    q.insert(a[i]);
  }
  vector<int> good(n + 1, 0);
  vector<int> rec(n, 0);
  for (int i = n - 1; i >= 0; i--) {
    int x = a[i];
    q.erase(x);
    auto it = q.rbegin();
    int cnt = 0;
    int el = -1;
    while (cnt < 2 && cnt < q.size() && *it > x) {
      cnt++;
      el = *it;
      it--;
    }
    if (!cnt) {
      rec[i] = 1;
    }
    if (cnt != 1) {
      continue;
    }
    good[el]++;
  }
  int maxk = -INF, pos = -1;
  for (int i = 0; i < n; i++) {
    if (good[a[i]] - rec[i] > maxk) {
      maxk = good[a[i]] - rec[i];
      pos = a[i];
    } else if (good[a[i]] - rec[i] == maxk) {
      pos = min(pos, a[i]);
    }
  }
  cout << pos;
}
