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 INF = 0x3f3f3f3f;
const long long int INFL = 0x3f3f3f3f3f3f3f3f;
const double pi = atan(1.0) * 4.0;
const int N = 1e5 + 500;
int v[N], ans[N];
set<int> st;
set<int>::iterator it;
int main(void) {
  ios_base::sync_with_stdio(false);
  int n;
  for (int i = 0; i < N; i++) ans[i] = 0;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> v[i];
  for (int i = 0; i < n; i++) {
    it = st.lower_bound(v[i]);
    if (it == st.end())
      ans[v[i]]--;
    else {
      int num = *it;
      ++it;
      if (it == st.end()) ans[num]++;
    }
    st.insert(v[i]);
  }
  int mx = -1;
  for (int i = 1; i <= n; i++) mx = max(mx, ans[i]);
  for (int i = 1; i <= n; i++) {
    if (ans[i] == mx) {
      cout << i << endl;
      return 0;
    }
  }
  return 0;
}
