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[100005];
int v[100005], n, i, mayor, lol, cosaMenor = 1000005;
int contMayor, numMenor;
bool record[100005];
int buenos;
void add(int pos) {
  while (pos > 0) {
    BIT[pos]++;
    pos -= pos & (-pos);
  }
}
int query(int pos) {
  int sum = 0;
  while (pos <= n) {
    sum += BIT[pos];
    pos += pos & (-pos);
  }
  return sum;
}
int cont[100005];
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &v[i]);
  for (i = 1; i <= n; i++) {
    if (query(v[i]) == 1)
      cont[mayor]++;
    else {
      if (query(v[i]) > 1) {
        cosaMenor = min(cosaMenor, v[i]);
      } else {
        record[v[i]] = true;
        buenos++;
      }
    }
    add(v[i]);
    mayor = max(mayor, v[i]);
  }
  for (i = 1; i <= n; i++) {
    if (buenos + cont[i] - (record[i] ? 1 : 0) > contMayor) {
      contMayor = buenos + cont[i] - (record[i] ? 1 : 0);
      numMenor = i;
    } else {
      if (buenos + cont[i] - (record[i] ? 1 : 0) == contMayor)
        numMenor = min(numMenor, i);
    }
  }
  if (contMayor == 0)
    printf("%d", cosaMenor > n ? 1 : cosaMenor);
  else
    printf("%d", numMenor);
  return 0;
}
