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
/* 
* @Author: steve
* @Date:   2015-04-01 21:42:19
* @Last Modified by:   steve
* @Last Modified time: 2015-04-01 21:49:02
*/
import java.io.*;
import java.util.*;

public class DZYLovesHash {
    public static void main(String[] args) throws Exception{
    	BufferedReader entrada = new BufferedReader(new InputStreamReader(System.in));
    	String[] cads= entrada.readLine().split(" ");
    	int p=Integer.parseInt(cads[0]), n=Integer.parseInt(cads[1]),j=-1;
    	boolean[] esta= new boolean[p];
    	for(int i=0;i<n;i++){
    		int numero=Integer.parseInt(entrada.readLine());
    		if(!esta[numero%p])
    			esta[numero%p]=true;
    		else{
    			j=i+1;
    			break;
    		}
    	}
    	System.out.println(j);
    }
}