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;
int n, pos = 1000000, ans = -1000000;
int a[100010], tr[100010], now[100010];
bool vis[100010];
inline void add(int x) {
  for (; x <= 100000; x += (x & -x)) {
    tr[x]++;
  }
}
inline int query(int x) {
  int ans = 0;
  for (; x; x -= (x & -x)) {
    ans += tr[x];
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  for (int i = 1; i <= n; ++i) {
    now[i] = query(a[i]);
    add(a[i]);
  }
  memset(tr, 0, sizeof(tr));
  for (int i = n; i >= 1; --i) {
    int tmp = query(a[i]);
    if (now[i] == i - 1) tmp--;
    if (tmp > ans) {
      ans = tmp;
      pos = a[i];
    } else if (tmp == ans && a[i] < pos) {
      pos = a[i];
    }
    if (now[i] == i - 2) add(a[i]);
  }
  cout << pos << endl;
  return 0;
}
