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.io.*;

public class Main {
	public static void main(String args[]) {new Main().run();}

	FastReader in = new FastReader();
	PrintWriter out = new PrintWriter(System.out);
	void run(){
		work();
		out.flush();
	}
	long mod=1000000007;
	long gcd(long a,long b) {
		return b==0?a:gcd(b,a%b);
	}
	void work() {
		int n=in.nextInt();
		if(n==1) {
			out.println(1);
			return;
		}
		int[] A=new int[n];
		for(int i=0;i<n;i++)A[i]=in.nextInt();
		int[] count=new int[n+1];
		for(int i=0,first=-99999999,second=-99999999;i<n;i++) {
			if(A[i]>first) {
				second=first;
				first=A[i];
				count[A[i]]=-1;
			}else if(A[i]>second) {
				count[first]++;
				second=A[i];
			}
		}
		
		int ret=1;
		int max=0;
		for(int i=1;i<=n;i++) {
			if(count[i]>count[ret]) {
				max=count[i];
				ret=i;
			}
		}
		out.println(ret);
	}
}



class FastReader
{
	BufferedReader br;
	StringTokenizer st;

	public FastReader()
	{
		br=new BufferedReader(new InputStreamReader(System.in));
	}

	public String next() 
	{
		if(st==null || !st.hasMoreElements())
		{
			try {
				st = new StringTokenizer(br.readLine());
			} catch (IOException e) {
				e.printStackTrace();
			}
		}
		return st.nextToken();
	}

	public int nextInt() 
	{
		return Integer.parseInt(next());
	}

	public long nextLong()
	{
		return Long.parseLong(next());
	}
}