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 long long MOD = 1e9 + 7;
const long long MAXN = 2e5 + 7;
long long a[MAXN];
bool good[MAXN];
long long want[MAXN], pref[MAXN], next_[MAXN];
signed main() {
  ios_base::sync_with_stdio(false);
  cout.precision(40);
  long long n;
  cin >> n;
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
  }
  stack<long long> st;
  for (long long i = 0; i < n; ++i) {
    while (st.size() && a[st.top()] < a[i]) {
      next_[st.top()] = i;
      st.pop();
    }
    st.push(i);
  }
  while (st.size()) {
    next_[st.top()] = n;
    st.pop();
  }
  long long mx1 = -MOD;
  long long mx2 = -MOD;
  for (long long i = 0; i < n; ++i) {
    if (a[i] < mx1 && a[i] > mx2) {
      want[i] = true;
    }
    if (a[i] > mx1) {
      mx2 = mx1;
      mx1 = a[i];
    } else if (a[i] > mx2) {
      mx2 = a[i];
    }
  }
  pref[0] = want[0];
  for (long long i = 1; i < n; ++i) {
    pref[i] = pref[i - 1] + want[i];
  }
  long long mx = -MOD;
  long long ans = -MOD;
  long long best = -1;
  for (long long i = 0; i < n; ++i) {
    mx = max(mx, a[i]);
    if (a[i] != mx) continue;
    long long j = next_[i];
    long long new_ans = pref[j - 1] - pref[i] - 1;
    if (new_ans > ans || (new_ans == ans && a[i] < a[best])) {
      ans = new_ans;
      best = i;
    }
  }
  mx = -MOD;
  for (long long i = 0; i < n; ++i) {
    mx = max(mx, a[i]);
    if (a[i] == mx) continue;
    long long new_ans = 0;
    if (new_ans > ans || (new_ans == ans && a[i] < a[best])) {
      ans = new_ans;
      best = i;
    }
  }
  cout << a[best] << '\n';
  return 0;
}
