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.util.*;

public class RemoveExtraOne {
    
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int[] p = new int[n];
        for (int i = 0; i < n; i++) {
            p[i] = in.nextInt();
        }
        
        int[] largest = new int[n];
        int[] secondLargest = new int[n];
        
        largest[0] = p[0];
        for (int i = 1; i < n; i++) {
            // is p[i] the second largest so far?
            if (secondLargest[i - 1] < p[i] && p[i] < largest[i - 1]) {
                secondLargest[i] = p[i];
                largest[i] = largest[i - 1];
            } else if (p[i] > largest[i - 1]) { // is p[i] the largest so far?
                secondLargest[i] = largest[i - 1];
                largest[i] = p[i];
            } else {
                secondLargest[i] = secondLargest[i - 1];
                largest[i] = largest[i - 1];
            }
        }
        
        int[] cntHaveAsLargestBefore = new int[n + 1];
        for (int i = 1; i < n; i++) {
            if (secondLargest[i - 1] < p[i] && p[i] < largest[i - 1]) {
                cntHaveAsLargestBefore[largest[i - 1]]++;
            }
        }
        //System.out.println(Arrays.toString(cntHaveAsLargestBefore));
        int max = cntHaveAsLargestBefore[p[0]] - 1;
        int ans = p[0];
        for (int i = 1; i < n; i++) {
            int newRecords = cntHaveAsLargestBefore[p[i]];
            if (p[i] > largest[i - 1]) {
                newRecords--;
            }
            if (newRecords > max) {
                max = newRecords;
                ans = p[i];
            } else if (newRecords == max) {
                ans = Math.min(ans, p[i]);
            }
        }
        System.out.println(ans);
        
        in.close();
        System.exit(0);
    }
}