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
/*
 * To change this template, choose Tools | Templates
 * and open the template in the editor.
 */


import java.io.*;
import java.util.*;
/**
 *
 * @author user
 */
public class Main {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args)throws IOException
    {
        // TODO code application logic here
        //BufferedReader in=new BufferedReader(new InputStreamReader(System.in));
        Scanner in=new Scanner(System.in);
        int p=in.nextInt();
        int n=in.nextInt();
        int a[]=new int[p+1];
        int b[]=new int[p+1];
        int count=0;
        for(int i=1;i<=n;i++)
        {
            int x=in.nextInt()%p;
            if(a[x]==0)
            {
                a[x]=1;
            }
            else
            {
                count=i;
                break;
            }
        }
        if(count>0)
            System.out.print(count);
        else
                System.out.print("-1");

    }

}
