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;
int n, AIB[100002];
int HowMany2nd[100002];
int nrrec[100002];
bool wasMax[100002];
int place[100002];
int max1;
void Add(int nod) {
  for (; nod <= n; nod += (nod & (-nod))) AIB[nod]++;
}
int Compute(int nod) {
  int s = 0;
  for (; nod; nod -= (nod & (-nod))) s += AIB[nod];
  return s;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int a;
    cin >> a;
    place[a] = i;
    if (a > max1)
      wasMax[a] = 1, nrrec[i] = nrrec[i - 1] + 1, max1 = a;
    else {
      nrrec[i] = nrrec[i - 1];
      if (Compute(max1) - Compute(a) == 1) ++HowMany2nd[max1];
    }
    Add(a);
  }
  int nrmax = nrrec[n];
  int who = 1;
  for (int i = 1; i <= n; ++i)
    if (!wasMax[i]) {
      who = i;
      break;
    }
  for (int i = 1; i <= n; ++i)
    if (wasMax[i] && nrrec[n] - 1 + HowMany2nd[i] > nrmax)
      nrmax = nrrec[n] - 1 + HowMany2nd[i], who = i;
  cout << who;
  return 0;
}
