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 inf = 1000000000;
long long mod = 1000000007LL;
long long mod2 = 998244353LL;
int t;
int n, m, k;
int a, b, c;
int p[100002];
int cnt[100002];
bool ok[100002];
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &p[i]);
  }
  set<int> ss;
  ss.clear();
  memset(cnt, 0, sizeof(cnt));
  memset(ok, false, sizeof(ok));
  int tot = 0;
  for (int i = 1; i <= n; ++i) {
    if (ss.empty()) {
      tot++;
      ok[i] = true;
    } else {
      auto itr = ss.upper_bound(p[i]);
      if (itr == ss.end()) {
        tot++;
        ok[i] = true;
      } else {
        auto itr2 = itr;
        itr2++;
        if (itr2 == ss.end()) {
          cnt[*itr]++;
        }
      }
    }
    ss.insert(p[i]);
  }
  int id = -1;
  int maxv = -1;
  for (int i = 1; i <= n; ++i) {
    if (ok[i]) {
      int res = tot - 1 + cnt[p[i]];
      if (res > maxv) {
        maxv = res;
        id = p[i];
      } else if (res == maxv && p[i] < id) {
        id = p[i];
      }
    } else {
      int res = tot + cnt[p[i]];
      if (res > maxv) {
        maxv = res;
        id = p[i];
      } else if (res == maxv && p[i] < id) {
        id = p[i];
      }
    }
  }
  cout << id << endl;
  return 0;
}
