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 + 7;
const int inf = INT_MAX;
const long long inff = 1e18;
const long long mod = 1e9 + 7;
int n;
int a[maxn], last[maxn];
vector<int> v, u;
vector<int> G[maxn];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  int maxx = -1, cnt = 0, L = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] > maxx)
      last[i] = L, L = i, maxx = a[i], u.push_back(i), cnt++;
    else
      v.push_back(a[i]);
  }
  maxx = -1;
  int ans = inf;
  for (int i : v) {
    ans = min(ans, i);
    maxx = cnt;
  }
  vector<int> tmp;
  a[0] = inf;
  tmp.push_back(0);
  for (int i = 1; i <= n; i++) {
    while (a[i] > a[tmp.back()]) tmp.pop_back();
    G[tmp.back()].push_back(a[i]);
    tmp.push_back(i);
  }
  a[0] = 0;
  for (int i : u) {
    sort((G[i]).begin(), (G[i]).end());
    int gi =
        ((int)(G[i]).size()) -
        (lower_bound((G[i]).begin(), (G[i]).end(), a[last[i]]) - G[i].begin());
    if (cnt - 1 + gi > maxx || cnt - 1 + gi == maxx && a[i] < ans) {
      maxx = cnt - 1 + gi;
      ans = a[i];
    }
  }
  cout << ans << endl;
}
