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;
const int N = 1e5 + 10;
int bits[N];
int num[N];
int n, m, T, k, p;
set<int> pt;
set<int>::iterator it;
void add(int i, int x) {
  while (i <= n) {
    bits[i] += x;
    i += (i & (-i));
  }
  return;
}
int sum(int i) {
  int res = 0;
  while (i > 0) {
    res += bits[i];
    i -= (i & (-i));
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p);
    if ((p - 1 > 0 && sum(p - 1) == i - 2) || (p - 1 == 0 && i == 2)) {
      it = pt.lower_bound(p);
      num[*it]++;
    }
    if ((p - 1 > 0 && sum(p - 1) == i - 1) || (p - 1 == 0 && i == 1)) num[p]--;
    pt.insert(p);
    add(p, 1);
  }
  int maxn = num[1];
  k = 1;
  for (int i = 2; i <= n; i++) {
    if (maxn < num[i]) {
      maxn = num[i];
      k = i;
    }
  }
  printf("%d\n", k);
  return 0;
}
