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 double PI = acos(-1.0);
const int INF = 1000 * 1000 * 1000 + 7;
const long long LINF = INF * (long long)INF;
const int MAX = 1000 * 100 + 47;
int P[MAX];
set<int> S;
int CNT[MAX];
int U[MAX];
int main() {
  int n;
  cin >> n;
  for (int i = (0); i < (n); i++) {
    cin >> P[i];
    P[i]--;
  }
  int cnt = 0;
  for (int i = (0); i < (n); i++) {
    set<int>::iterator it = S.upper_bound(P[i]);
    if (it == S.end()) {
      U[P[i]] = 1;
      cnt++;
      S.insert(P[i]);
      continue;
    }
    int a = *it;
    it++;
    if (it != S.end()) {
      S.insert(P[i]);
      continue;
    }
    S.insert(P[i]);
    CNT[a]++;
  }
  int ans = -INF;
  int elem = -1;
  for (int i = (0); i < (n); i++) {
    int res = cnt;
    if (U[i]) res--;
    res += CNT[i];
    if (ans < res) {
      ans = res;
      elem = i;
    }
  }
  cout << elem + 1 << endl;
}
