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() {
  long n, i, b1 = 0, b2 = 0, t, wa = -1, nwa = 2147483647L, tmp = -1,
             nwa2 = 2147483647L;
  scanf("%ld", &n);
  scanf("%ld", &t);
  if (n == 1) {
    printf("%ld\n", t);
    return 0;
  }
  tmp = -1;
  b1 = t;
  for (i = 1; i < n; i++) {
    scanf("%ld", &t);
    if (t > b1) {
      if (tmp > wa || (tmp == wa && b1 < nwa)) {
        wa = tmp;
        nwa = b1;
      }
      tmp = -1;
      b2 = b1;
      b1 = t;
    } else if (t > b2) {
      tmp++;
      b2 = t;
    } else {
      if (wa < 0) {
        nwa = t;
        wa = 0;
      }
      if (wa == 0 && t < nwa2) nwa2 = t;
    }
  }
  if (tmp > wa || (tmp == wa && b1 < nwa)) {
    nwa = b1;
    wa = tmp;
  }
  if (wa != 0)
    printf("%ld\n", nwa);
  else
    printf("%ld\n", nwa2);
}
