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.*;
import java.util.Map.Entry;
import java.io.*;
public class arcA {
	public static void main(String[] args) {
		MyScanner sc=new MyScanner();
		PrintWriter pw=new PrintWriter(System.out);
		
		int n=sc.nextInt();
		int [] a=new int [n];
		for(int i=0;i<n;i++)
			a[i]=sc.nextInt();
		TreeSet<Integer> set=new TreeSet<Integer>();
		int [] c=new int [n+1];
		for(int i=0;i<n;i++){
			if(set.isEmpty())
				c[a[i]]--;
			else
				if(a[i]>set.last())
					c[a[i]]--;
				else
					if(set.last()>a[i]){
						int x=set.higher(a[i]);
						if(x==set.last())
							c[x]++;
					}
			set.add(a[i]);
		}
//		System.err.println(Arrays.toString(c));
		int ans=-1;
		int Addval=-5;
		for(int i=1;i<=n;i++)
			if(c[i]>Addval){
				Addval=c[i];
				ans=i;
			}
		
		
		pw.println(ans);
		pw.flush();
		pw.close();
	}
	
	static class MyScanner{
		BufferedReader br;
		StringTokenizer st;
		
		MyScanner(){
			br=new BufferedReader(new InputStreamReader(System.in));
		}
		String next(){
			while(st==null || !st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (Exception e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		int nextInt(){
			return Integer.parseInt(next());
		}
		long nextLong(){
			return Long.parseLong(next());
		}
	}
}
