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;
long long a[100005], cnt[100005] = {0};
int main() {
  ios_base::sync_with_stdio(0);
  long long n, ans, pos, mx = 0, smx = 0;
  cin >> n;
  for (int i = 0; i < n; cin >> a[i++])
    ;
  for (int i = 0; i < n; ++i) {
    if (a[i] > mx) {
      if (i == 1 && mx == a[0]) --cnt[mx];
      smx = mx;
      mx = a[i];
      if (i > 0) --cnt[mx];
    } else if (a[i] > smx) {
      if (i > 1 || mx != a[0]) ++cnt[mx];
      smx = a[i];
    }
  }
  ans = cnt[a[0]], pos = a[0];
  for (int i = 1; i < n; ++i)
    if (ans < cnt[a[i]] || (ans == cnt[a[i]] && a[i] < pos))
      ans = cnt[a[i]], pos = a[i];
  cout << pos;
}
