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 A 
{
    
    public static void main(String[] args) {

        int n , p;
        boolean[] mod=new boolean[300+10];
        Scanner sc=new Scanner(System.in);
        boolean flag=false;

        p=sc.nextInt();
        n=sc.nextInt();
        
        for(int i=1 ; i<=n ; i++)
        {   
            int x=sc.nextInt();
        
            if(flag)
                continue;
            
            if(mod[x%p])
            {
                System.out.println(i);
                flag=true;
            }
            
            else
                mod[x%p]=true;
        }
        
        if(!flag)
            System.out.println(-1);
        
        sc.close();
    }
}
