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, c[100005], p[100005], f[100005];
int sum(int x) {
  int ans = 0;
  for (int i = x; i > 0; i -= i & (-i)) ans += c[i];
  return ans;
}
void add(int x, int w) {
  for (int i = x; i <= n; i += i & (-i)) c[i] += w;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &p[i]);
  for (int i = 1; i <= n; ++i) {
    f[i] = sum(p[i]);
    add(p[i], 1);
  }
  int ans = -100, id = 1e9;
  for (int i = 1; i <= n; ++i) c[i] = 0;
  for (int i = n; i >= 1; --i) {
    int tmp = sum(p[i]);
    if (f[i] == i - 1) tmp--;
    if ((tmp > ans) || (tmp == ans && p[i] < id)) {
      ans = tmp;
      id = p[i];
    }
    if (f[i] == i - 2) add(p[i], 1);
  }
  printf("%d\n", id);
  return 0;
}
