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 N = 2e6 + 10;
const long long mod = 998244353;
long long power(long long x, long long y) {
  x = x % mod, y = y % (mod - 1);
  long long ans = 1;
  while (y > 0) {
    if (y & 1) {
      ans = (1ll * x * ans) % mod;
    }
    y >>= 1;
    x = (1ll * x * x) % mod;
  }
  return ans;
}
long long add(long long a, long long b) { return (a % mod + b % mod) % mod; }
long long mul(long long a, long long b) { return (a % mod * b % mod) % mod; }
long long n, k;
const long long K = 210;
vector<long long> a(N);
vector<long long> b(N);
set<long long> st;
int32_t main() {
  long long n;
  cin >> n;
  a.resize(n + 1);
  map<long long, long long> mp;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    st.insert(a[i]);
    auto it = st.upper_bound(a[i]);
    if (it == st.end())
      mp[a[i]]--;
    else if (it == (--st.end()))
      mp[*it]++;
  }
  long long ans = INT_MIN, temp = 0;
  for (long long i = 1; i <= n; i++) {
    if (mp[i] > ans) {
      ans = mp[i];
      temp = i;
    }
  }
  cout << temp;
  return 0;
}
