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 tong[100000 + 10];
int a[100000 + 10];
int no[100000 + 10];
set<int> st;
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  st.insert(a[1]);
  tong[a[1]]--;
  for (int i = 2; i <= n; i++) {
    int save = (*st.rbegin());
    if (a[i] < save) {
      st.erase(save);
      if (st.empty() || a[i] > (*st.rbegin())) tong[save]++;
      st.insert(save);
    } else
      tong[a[i]]--;
    st.insert(a[i]);
  }
  int ans = -1000000, p = 1000000;
  for (int i = 1; i <= n; i++)
    if (ans < tong[i]) ans = max(ans, tong[i]), p = i;
  cout << p << '\n';
  return 0;
}
