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;
template <class T>
void out(T A[], int n) {
  for (int i = 0; i < n; i++) cout << A[i] << " ";
  cout << endl;
}
template <class T>
void out(vector<T> A, int n = -1) {
  if (n == -1) n = A.size();
  for (int i = 0; i < n; i++) cout << A[i] << " ";
  cout << endl;
}
const int N = 100000 + 5;
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
int n;
int a[N], cnt[N];
pair<int, int> c[N];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
  }
  int fir, sec;
  fir = sec = -INF;
  memset(cnt, 0, sizeof(cnt));
  for (int i = 0; i < n; ++i) {
    if (a[i] > fir) {
      --cnt[a[i]];
      sec = fir;
      fir = a[i];
    } else if (a[i] > sec) {
      ++cnt[fir];
      sec = a[i];
    }
  }
  int max_cnt = -INF;
  int ans = -1;
  for (int i = 1; i <= n; ++i) {
    if (cnt[i] > max_cnt) {
      max_cnt = cnt[i];
      ans = i;
    } else if (cnt[i] == max_cnt) {
      ans = min(ans, i);
    }
  }
  printf("%d\n", ans);
  return 0;
}
