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;
template <class T>
bool umin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool umax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
int F[100009], arr[100009], ans[100009], imp[100009], pos[100009];
void upd(int x, int val) {
  for (; x; x -= x & (-x)) F[x] += val;
}
int tap(int x) {
  int res = 0;
  for (; x < 100009; x += x & (-x)) res += F[x];
  return res;
}
set<int> s;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", arr + i), pos[arr[i]] = i;
  int all = 0;
  for (int i = 1; i <= n; i++) {
    int res = tap(arr[i]);
    if (!res)
      imp[i] = 1;
    else if (res == 1) {
      int val = *s.rbegin();
      ans[pos[val]]++;
    }
    s.insert(arr[i]);
    upd(arr[i], +1);
    all += imp[i];
  }
  int mx = -1, who;
  for (int i = 1; i <= n; i++) {
    int val = all - imp[i] + ans[i];
    if (mx < val)
      mx = val, who = arr[i];
    else if (mx == val)
      umin(who, arr[i]);
  }
  printf("%d\n", who);
  return 0;
}
