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 MAXN = 100005;
const int INF = 1 << 29;
set<pair<int, int> > myset;
int a[MAXN], v[MAXN], act[MAXN];
int n;
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  act[1] = 1;
  myset.insert(make_pair(a[1], 1));
  myset.insert(make_pair(-INF, 0));
  for (int i = 2; i <= n; i++) {
    set<pair<int, int> >::reverse_iterator x = myset.rbegin();
    if (a[i] > x->first)
      act[i] = 1;
    else {
      set<pair<int, int> >::reverse_iterator y = x;
      ++y;
      if (a[i] > y->first) ++v[x->second];
    }
    myset.insert(make_pair(a[i], i));
  }
  int pos = 0, val = -INF;
  for (int i = 1; i <= n; i++) {
    if (v[i] - act[i] > val || (v[i] - act[i] == val && a[i] < a[pos])) {
      val = v[i] - act[i];
      pos = i;
    }
  }
  cout << a[pos] << endl;
  return 0;
}
