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, a[100005], tut[100005], cev[100005], mx[100005], ans = -1,
                                                        yaz = 1000000009;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  stack<pair<int, int> > s;
  s.push(make_pair(1000000009, 0));
  for (int i = 1; i <= n; i++) {
    mx[i] = max(mx[i - 1], a[i]);
    if (mx[i] == a[i]) cev[i]--;
    while (a[i] > s.top().first) s.pop();
    tut[i] = s.top().second;
    s.push(make_pair(a[i], i));
  }
  for (int i = 1; i <= n; i++) {
    int go = tut[i];
    if (mx[go - 1] < a[i]) cev[go]++;
  }
  for (int i = 1; i <= n; i++) {
    if (cev[i] > ans) {
      ans = cev[i];
      yaz = a[i];
    } else if (cev[i] == ans) {
      yaz = min(yaz, a[i]);
    }
  }
  printf("%d", yaz);
}
