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 bit[100100], n;
inline void add(int pos, int val) {
  while (pos <= n) {
    bit[pos] += val;
    pos += pos & (-pos);
  }
}
inline int sum(int pos) {
  int ret = 0;
  while (pos > 0) {
    ret += bit[pos];
    pos -= pos & (-pos);
  }
  return ret;
}
int ans = -0x3f3f3f3f, id = 100100;
int a[100100];
int ax[100100];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    int num = sum(n) - sum(a[i] - 1);
    if (num > 1) {
      add(a[i], 1);
      continue;
    }
    if (num == 0) {
      add(a[i], 1);
      --ax[a[i]];
      continue;
    }
    int lb = a[i], ub = n;
    while (ub > lb + 1) {
      int mid = (lb + ub) >> 1;
      if (sum(mid) > sum(a[i]))
        ub = mid;
      else
        lb = mid;
    }
    add(a[i], 1);
    ++ax[ub];
  }
  for (int i = 1; i <= n; ++i)
    if (ax[i] > ans) {
      ans = ax[i];
      id = i;
    }
  printf("%d\n", id);
  return 0;
}
