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.io.*;

public class Main {

    public static void main(String[] args)throws IOException {
        // TODO Auto-generated method stub

        BufferedReader rd=new BufferedReader(new InputStreamReader(System.in));
        int p,n,num;
        boolean flag=false;
        String input=rd.readLine();
        String inp[]=input.split(" ");
        p=Integer.parseInt(inp[0]);
        n=Integer.parseInt(inp[1]);
        
        int arr[]=new int[p];
        
        for(int j=0;j<p;++j)
            arr[j]=-1;
        
        for(int i=0;i<n;++i)
        {
            num=Integer.parseInt(rd.readLine());
            if(arr[num%p]==-1)
                arr[num%p]=num;
            else
            {   
                System.out.println(i+1);
                flag=true;
                break;
            }
            
        }
        
        if(!flag)
            System.out.println("-1");
    }

}
