DZY has a hash table with p buckets, numbered from 0 to p - 1. He wants to insert n numbers, in the order they are given, into the hash table. For the i-th number xi, DZY will put it into the bucket numbered h(xi), where h(x) is the hash function. In this problem we will assume, that h(x) = x mod p. Operation a mod b denotes taking a remainder after division a by b.

However, each bucket can contain no more than one element. If DZY wants to insert an number into a bucket which is already filled, we say a "conflict" happens. Suppose the first conflict happens right after the i-th insertion, you should output i. If no conflict happens, just output -1.

Input

The first line contains two integers, p and n (2 ≤ p, n ≤ 300). Then n lines follow. The i-th of them contains an integer xi (0 ≤ xi ≤ 109).

Output

Output a single integer — the answer to the problem.

Examples

Input

10 5
0
21
53
41
53


Output

4


Input

5 5
0
1
2
3
4


Output

-1
import java.util.Scanner;


public class DZY_Loves_Hash {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
        Scanner in=new Scanner (System.in);
        int p=in.nextInt();
        int n=in.nextInt();
        int arr[]=new int[n];
        boolean check[]=new boolean[p];
        int z=0;
        boolean t=false;
        for(int i=0;i<n;i++){
        	int k=in.nextInt();
        	z=(k%p);
        //	System.out.println(z);
        
        	if(check[z]==true)
        	{
        		System.out.println(i+1);
        		t=true;
        		return;
        		
        	}
        	
        	check[z]=true;
        
}
        if(t==false)
        	System.out.println(-1);
}

}
