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, sum[1000001], nums[1000001], p[1000001];
void update(int p) {
  while (p <= n) {
    sum[p]++;
    p = p | (p + 1);
  }
}
int interval_sum(int p) {
  int result = 0;
  while (p >= 0) {
    result += sum[p];
    p = (p & (p + 1)) - 1;
  }
  return result;
}
int main() {
  scanf("%d", &n);
  if (n == 1) {
    printf("1");
    return 0;
  }
  int max_value = 0, index = 0;
  int min_value = n + 1;
  for (int i = 0; i < n; i++) {
    scanf("%d", &nums[i]);
    int res = i - interval_sum(nums[i] - 1);
    if (res == 1) {
      p[index]++;
      min_value = min(min_value, nums[i]);
    } else if (res == 0) {
      p[i] = -1;
    } else {
      min_value = min(min_value, nums[i]);
    }
    if (nums[i] > max_value) {
      max_value = nums[i];
      index = i;
    }
    update(nums[i]);
  }
  int extra = 1000001 + 1, freq = 0;
  for (int i = 0; i < n; i++) {
    if (p[i] > freq) {
      freq = p[i];
      extra = nums[i];
    } else if (p[i] == freq) {
      extra = min(extra, nums[i]);
    }
  }
  if (freq <= 0) {
    extra = min_value;
    if (min_value == n + 1) extra = 1;
  }
  printf("%d \n", extra);
  return 0;
}
