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.
//Codeforces 900.3
import java.util.*;

public class Remove {
	public static int numRecord(int[] p) {
		if (p.length == 1) return 0;
		int max = p[0], ans = 0;
		for (int i = 1; i < p.length; i++) if (p[i] > max) {
			ans++;
			max = p[i];
		}
		return ans;
	}
	public static void main(String[] args) {
		Scanner f = new Scanner(System.in);
		int n = f.nextInt(), max = 0;
		int[] values = new int[n];
		boolean[] record = new boolean[n];
		for (int i = 0; i < n; i++) {
			values[i] = f.nextInt();
			if (values[i] > max) {
				record[i] = true;
				max = values[i];
			}
		}
		//We have an O(N) solution: for each record, we consider the array between that number and the next record, w/ the starting number being the previous record, and count.
		int ans = -1, benefit = -1, prev = 0, i = 0;
		while (i < record.length) {
			int start = i;
			//Get to the next instance
			i++;
			while (i < record.length && !record[i]) i++;
			if (start == 0) {
				int[] good = Arrays.copyOfRange(values, start, i);
				good[0] = 0;
				ans = start;
				benefit = numRecord(good) - 1;
//				System.out.println(start + " " + benefit);
			}
			else {
				int[] good = Arrays.copyOfRange(values, start, i);
				good[0] = values[prev];
				int b = numRecord(good) - 1;
				if (b > benefit || (b == benefit && values[start] < values[ans])) {
					ans = start;
					benefit = b;
				}
//				System.out.println(i + " " + b);
			}
			prev = start;
		}
		if (benefit <= 0) {		//We compare our answer to non-peaks and find the smallest.
			for (i = 0; i < n; i++) {
				if (record[i]) continue;
				int b = 0;
				if (b > benefit || (b == benefit && values[i] < values[ans])) {
					ans = i;
					benefit = b;
				}
			}
		}
		System.out.println(values[ans]);
	}
}
