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;
class fenwick {
  int n;
  vector<int> tree;

 public:
  fenwick(int _n) : n(_n), tree(n + 5, 0) {}
  void Update(int pos, int val, int limit) {
    while (pos <= limit) {
      tree[pos] += val;
      pos += pos & (-pos);
    }
  }
  int Query(int pos) {
    int sum = 0;
    while (pos > 0) {
      sum += tree[pos];
      pos -= pos & (-pos);
    }
    return sum;
  }
};
void solve() {
  int n;
  cin >> n;
  vector<int> v(n);
  for (int& in : v) cin >> in;
  vector<int> mark(n + 1, -1);
  set<int> st(v.begin(), v.end());
  fenwick fen(n);
  int mx = 0;
  map<int, int> mp;
  for (int i = 0; i < n; i++) {
    if (fen.Query(n) - fen.Query(v[i]) == 0)
      st.erase(v[i]);
    else if (fen.Query(n) - fen.Query(v[i]) == 1)
      mp[mx]++;
    fen.Update(v[i], 1, n);
    mx = max(mx, v[i]);
  }
  if (mp.size()) {
    int val = 0, mxx = -1;
    for (auto it : mp)
      if (it.second > mxx) val = it.first, mxx = it.second;
    if (mxx == 1 && st.size())
      cout << min(val, *st.begin()) << "\n";
    else
      cout << val << "\n";
  } else if (st.size())
    cout << *st.begin() << "\n";
  else
    cout << 1 << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int tc = 1;
  while (tc--) {
    solve();
  }
  return 0;
}
