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 <typename T>
constexpr T sqr(T x) {
  return x * x;
}
constexpr unsigned long long bit(int num) { return 1ull << num; }
int main() {
  int n;
  cin >> n;
  vector<int> v(n);
  vector<int> pos(n);
  for (int i = 0; i < int(n); ++i) {
    cin >> v[i];
    --v[i];
    pos[v[i]] = i;
  }
  vector<int> cnt(n);
  int min1 = n;
  int i1 = -1;
  int min2 = n;
  int i2 = -1;
  for (int i = int(n) - 1; i >= 0; --i) {
    if (pos[i] > min2) continue;
    if (pos[i] > min1) {
      min2 = pos[i];
      i2 = i;
      ++cnt[i1];
    } else {
      min2 = min1;
      i2 = i1;
      min1 = pos[i];
      i1 = i;
    }
  }
  int curmax = -1;
  for (int i = 0; i < int(n); ++i) {
    if (v[i] > curmax) {
      if (cnt[v[i]] == 0) --cnt[v[i]];
      curmax = v[i];
    }
  }
  for (int i = 0; i < int(n); ++i)
    if (cnt[i] == 1) cnt[i] = 0;
  auto res = max_element((cnt).begin(), (cnt).end()) - cnt.begin() + 1;
  cout << res << endl;
  return 0;
}
