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 MAXN = 1e5 + 7;
int n;
int a[MAXN], r[MAXN], x[MAXN], vt[MAXN];
set<int> tset;
int main() {
  cin >> n;
  int tmax = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    vt[a[i]] = i;
    if (a[i] > tmax) {
      r[i] = 1;
      tmax = a[i];
    }
  }
  tset.insert(a[0]);
  x[0] -= r[0];
  for (int j = 1; j < n; j++) {
    tset.insert(a[j]);
    set<int>::iterator it = tset.end();
    it--;
    it--;
    set<int>::iterator it2 = tset.end();
    it2--;
    if (*it == a[j]) {
      x[vt[*it2]]++;
    }
    x[j] -= r[j];
  }
  int ans = a[0];
  tmax = x[0];
  for (int i = 0; i < n; i++) {
    if (x[i] > tmax) {
      tmax = x[i];
      ans = a[i];
    } else if (x[i] == tmax) {
      ans = min(ans, a[i]);
    }
  }
  cout << ans;
}
