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.
//package leto;

import java.io.FileNotFoundException;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;

public class math {
	static int num = 1;
	static int cons = 1000000;
	public static void main(String[] args) throws FileNotFoundException {
		Scanner scan = new Scanner(System.in);
	/*	int n = scan.nextInt();
		int a = scan.nextInt();
		int b = scan.nextInt();
		int c = scan.nextInt();
		int [][][]d = new int [a+1][b+1][c+1];
		d[0][0][0] = n;
		for(int i = 0;i<=a;i++){
			for(int j = 0;j<=b;j++){
				for(int k = 0;k<=c;k++){
					
				}
			}
		}*/
		int n = scan.nextInt();
		int []a = new int [n];
		if(n == 1){
			System.out.println(1);
			return;
		}
		int max1 = -1;
		int max2 = -1;
		int []count = new int [n];
		
		for(int i = 0;i<n;i++){
			a[i] = scan.nextInt();
			if((max2 == -1 || a[i] > a[max2]) && (max1 != -1 && a[i] < a[max1])){
				count[max1]++;
			}
			if(max1 == -1 || a[i] > a[max1]){
				count[i]--;
			}
			if(max1 == -1){
				max1 = i;
				continue;
			}
			if(a[i] >= a[max1]){
				max2 = max1;
				max1 = i;
			} else {
				if(max2 == -1){
					max2 = i;
					continue;
				}
				if(a[i] >= a[max2]){
					max2 = i;
				}
			}
		}
		int ans = 0;
		int max = count[0];
		for(int i = 0;i<n;i++){
			if(count[i] > max || (count[i] == max && a[i] < a[ans]) ){
				max = count[i];
				ans = i;
			}
		}
		System.out.println(a[ans]);
		/*boolean []visit = new boolean [cons];
		int a = scan.nextInt();
		int b = scan.nextInt();
		int c = scan.nextInt();
		int count = 0;
		int pos = -1;
		while(!visit[a]){
			count++;
			visit[a] = true;
			a *= 10;
			if(a/b == c){
				pos = count;
				break;
			} else {
				a %= b;
			}
		}
		System.out.println(pos);*/
	}
}
