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;
long long fpow(long long n, long long k, long long p = 1000000007) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
long long inv(long long a, long long p = 1000000007) {
  return fpow(a, p - 2, p);
}
int main() {
  long long n;
  cin >> n;
  long long a[100003];
  long long cnt[100003] = {0};
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long maxx = -1;
  long long maxx2 = -1;
  for (long long i = 0; i < n; i++) {
    if (i == 0) {
      cnt[a[i]] = -1;
      maxx = a[i];
    } else {
      if (a[i] > maxx) {
        cnt[a[i]] = -1;
        maxx = a[i];
      }
    }
  }
  maxx = -1;
  long long maxxans = 0;
  for (long long i = 0; i < n; i++) {
    if (a[i] > maxx) {
      maxx2 = maxx;
      maxx = a[i];
    } else if (a[i] > maxx2) {
      maxx2 = a[i];
      cnt[maxx]++;
      if (cnt[maxx] > maxxans) {
        maxxans = cnt[maxx];
      }
    }
  }
  for (long long i = 1; i <= n; i++) {
    if (cnt[i] == maxxans) {
      cout << i << "\n";
      return 0;
    }
  }
  maxxans = -1;
  for (long long i = 1; i <= n; i++) {
    if (cnt[i] == maxxans) {
      cout << i << "\n";
      break;
    }
  }
}
