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 int maxn = 1e6 + 1;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<bool> r(n, 0);
  int a[n];
  for (auto(i) = (0); (i) < (n); ++i) {
    cin >> a[i];
  }
  int maxa = -1;
  for (auto(i) = (0); (i) < (n); ++i) {
    if (a[i] > maxa) {
      r[i] = 1;
      maxa = a[i];
    }
  }
  vector<int> k(n, 0);
  int maxu = -1;
  for (auto(i) = (0); (i) < (n); ++i) {
    if (i) {
      maxu = max(a[i - 1], maxu);
    }
    if (!r[i]) continue;
    maxa = maxu;
    int counter = 0;
    for (auto(j) = (i + 1); (j) < (n); ++j) {
      maxa = max(maxa, a[j]);
      if (r[j]) {
        k[i] = counter;
        break;
      } else if (a[j] >= maxa)
        counter++;
    }
    k[i] = counter;
  }
  int maxd = -1, pos = -1, mina = 1e9 + 7, nr = -1;
  for (auto(i) = (0); (i) < (n); ++i) {
    if (maxd < k[i]) {
      pos = i;
      maxd = k[i];
    }
    if (maxd == k[i]) {
      if (a[pos] > a[i]) {
        pos = i;
      }
    }
    if (!r[i] && mina > a[i]) {
      mina = a[i];
      nr = i;
    }
  }
  if (maxd <= 1) {
    if (mina == 1e9 + 7) {
      cout << 1;
      return 0;
    }
    cout << mina;
    return 0;
  }
  cout << a[pos];
  return 0;
}
