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 mxsz = 1e5 + 5;
int sz, arr[mxsz], benefit[mxsz], records[mxsz];
int solve() {
  set<int> cur;
  set<int>::iterator it;
  for (int i = 0; i < sz; i++) {
    it = cur.upper_bound(arr[i]);
    if (it == cur.end())
      records[arr[i]] = 1;
    else if ((++it) == cur.end())
      benefit[*(--it)]++;
    cur.insert(arr[i]);
  }
  int sol = -1e9, ret = 1;
  for (int i = sz; i >= 1; i--) {
    if (benefit[i] - records[i] >= sol) {
      sol = benefit[i] - records[i];
      ret = i;
    }
  }
  return ret;
}
int main() {
  scanf("%d", &sz);
  for (int i = 0; i < sz; i++) scanf("%d", &arr[i]);
  printf("%d\n", solve());
  return 0;
}
