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 main() {
  int n, p[100010], i, arec = 0, cnt[100010], maxi = -1, sec_max = -1;
  scanf("%d", &n);
  int marked[100010];
  scanf("%d", &p[0]);
  marked[0] = 1;
  maxi = p[0];
  arec++;
  for (i = 1; i < n; i++) {
    scanf("%d", &p[i]);
    if (maxi < p[i])
      arec++, marked[i] = 1;
    else if (sec_max < p[i])
      sec_max = p[i], cnt[maxi]++;
    if (maxi <= p[i])
      sec_max = maxi, maxi = p[i];
    else if (sec_max == -1)
      sec_max = p[i];
  }
  int ans = -1, bit = -1;
  for (i = 0; i < n; i++) {
    int temp = arec - marked[i] + cnt[p[i]];
    if (bit < temp)
      bit = temp, ans = p[i];
    else if (bit == temp && ans > p[i])
      ans = p[i];
  }
  printf("%d\n", ans);
  return 0;
}
