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.*;
public class Mian {
	public static void main (String[] args) {
		Scanner sc = new Scanner(System.in);
		boolean arr[] = new boolean [sc.nextInt()],flag = true;
		int n = sc.nextInt();
		for(int i=1 ; i<=n ; i++){
		    int mem = sc.nextInt();
		    if(!arr[mem%arr.length]){
		        arr[mem%arr.length] = true;
		    }else{
		        flag = false;
		        System.out.println(i);
		        break;
		    }
		}
		if(flag)
		    System.out.println("-1");
	}
}