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.*;
import java.util.*;

public class C {

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb = new StringBuilder();
        int n = Integer.parseInt(br.readLine());
        StringTokenizer st = new StringTokenizer(br.readLine());
        int[] arr = new int[n];
        for(int i = 0; i<n; i++){
            arr[i] = Integer.parseInt(st.nextToken());
        }
        ArrayList<Integer> max_ndx = new ArrayList<>();
        int curr_max = 0;
       for(int i = 0; i<n; i++){
           if(arr[i]>curr_max){
               max_ndx.add(i);
               curr_max = arr[i];
           }
       }
       max_ndx.add(n);
        int ans = n+1, max_release = 0;
       if(arr.length>1) {
           for (int i = 1; i < max_ndx.size(); i++) {
               int release = -1;
               int s = max_ndx.get(i - 1) + 1, e = max_ndx.get(i) - 1;
               if(s>e) continue;
               int max = arr[s], min = arr[s];
               for (int k = s; k <= e; k++) {
                   if (arr[k] >= max) {
                       if (i == 1 || arr[k] > arr[max_ndx.get(i - 2)]) release++;
                       max = arr[k];
                   }
                   if(min>arr[k]) min = arr[k];
               }
               if (max_release < release) {
                   ans = arr[s - 1];
                   max_release = release;
               }
               if(max_release == 0 && min<ans){
                   ans = min;
               }
           }
           System.out.println((ans == n+1)?1:ans);
       }
       else{
           System.out.println(1);
       }
    }
}
