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.
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.Scanner;
import java.util.stream.IntStream;

public class C {

	public static void main(String[] args) {
		try (Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
				PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)))) {
			final int n = in.nextInt();
			final int[] a = new int[n];
			for (int i = 0; i < n; i++) {
				a[i] = in.nextInt();
			}
			out.println(solve(a));
		}
	}

	private static int solve(int[] a) {
		final int n = a.length;
		final int[] ans = new int[n + 1];
		int m1 = 0, m2 = 0;
		for (int i = 0; i < n; i++) {
			if (a[i] > m1) {
				m2 = m1;
				m1 = a[i];
				ans[m1]--;
			} else if (a[i] > m2) {
				m2 = a[i];
				ans[m1]++;
			}
		}
		return IntStream.range(1, n + 1).reduce(1, (i, j) -> ans[i] < ans[j] ? j : i);
	}

}
