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 gcd(long long x, long long y) {
  if (y == 0)
    return x;
  else
    return gcd(y, x % y);
}
long long expo(long long n, long long m, long long p) {
  long long r = 1;
  while (m > 0) {
    if (m % 2) r = (r * n) % p;
    n = (n * n) % p;
    m = m / 2;
  }
  return r % p;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  set<long long> s;
  vector<long long> x(n + 1);
  x[a[0]]--;
  s.insert(-a[0]);
  for (long long i = 1; i < n; i++) {
    long long p = -1 * (*s.begin());
    long long P = p;
    if (p < a[i])
      x[a[i]]--;
    else {
      if (s.size() == 1) {
        x[p]++;
        s.insert(-a[i]);
        continue;
      }
      auto it = s.begin();
      it++;
      p = -1 * (*it);
      if (p < a[i]) x[P]++;
    }
    s.insert(-a[i]);
  }
  long long ans = 0, m = -10000000000000000;
  for (long long i = 1; i <= n; i++) {
    if (m < x[i]) {
      ans = i;
      m = x[i];
    }
  }
  cout << ans;
}
