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 N = 1e5 + 1, INF = 1e7;
const double PI = 3.141592653589793238462643383279502884197, eps = 1e-6;
int num[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, x, mx, ans = 1;
  set<int> st, sum;
  set<int>::iterator it1, it2;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> x;
    if (st.size() != 0) {
      it1 = it2 = st.end();
      it1--, it2--, it2--;
      if (*it1 > x && ((st.size() > 1 && *it2 < x) || st.size() == 1))
        num[*it1]++;
      else if (*it1 < x)
        sum.insert(x);
    } else
      sum.insert(x);
    st.insert(x);
  }
  mx = sum.size() - (sum.find(1) != sum.end());
  for (int i = 2; i <= n; i++) {
    num[i] += sum.size() - (sum.find(i) != sum.end());
    if (num[i] > mx) mx = num[i], ans = i;
  }
  cout << ans;
  return 0;
}
