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 a[100005] = {0}, b[100005] = {0}, c[100005] = {0};
int main() {
  int n, i, max1, max2, t1 = 0, ans;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d", &a[i]);
  if (n <= 2)
    printf("%d\n", a[0]);
  else {
    max1 = min(a[0], a[1]);
    max2 = max(a[0], a[1]);
    c[a[0]]++;
    if (a[0] < a[1])
      c[a[1]]++;
    else
      b[a[0]]++;
    for (i = 2; i < n; i++) {
      if (a[i] > max2) {
        c[a[i]]++;
        max1 = max2;
        max2 = a[i];
      } else if (a[i] > max1) {
        b[max2]++;
        max1 = a[i];
      }
    }
    for (i = 1; i <= n; i++)
      if (c[i] > 0) t1++;
    ans = a[0];
    for (i = 1; i < n; i++)
      if (t1 - c[a[i]] + b[a[i]] > t1 - c[ans] + b[ans])
        ans = a[i];
      else if (t1 - c[a[i]] + b[a[i]] == t1 - c[ans] + b[ans] && a[i] < ans)
        ans = a[i];
    printf("%d\n", ans);
  }
  return 0;
}
