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 base = 1;
int tree[500000];
int ile[500000];
int tab[500000];
void insert(int a, int c) {
  a += base;
  while (a >= 1) {
    tree[a] += c;
    a = a >> 1;
  }
}
int query(int a, int b) {
  if (b <= a) return 0;
  a += base;
  b += base;
  int res = 0;
  while (a / 2 != b / 2) {
    if (a % 2 == 0) res += tree[a + 1];
    if (b % 2 == 1) res += tree[b - 1];
    a = a >> 1;
    b = b >> 1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  while (base < n + 2) base = base << 1;
  int res = 0;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    tab[i] = x;
    ile[i] = query(x, n + 1);
    if (ile == 0) {
      res++;
    }
    insert(x, 1);
  }
  for (int i = 1; i <= 2 * base + 1; i++) tree[i] = 0;
  int wynik = -2;
  int odp = -1;
  for (int i = n; i >= 1; i--) {
    if (ile[i] == 0) res--;
    int kandydat = res + query(0, tab[i]);
    if (kandydat > wynik || (kandydat == wynik && tab[i] < odp)) {
      wynik = kandydat;
      odp = tab[i];
    }
    if (ile[i] == 1) insert(tab[i], 1);
    if (ile[i] == 0) res++;
  }
  cout << odp << endl;
}
