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 dzyloveshash{
	public static void main(String[] args){
		Scanner in = new Scanner(System.in);
		int p, n, i, inc = 1, flag = 0;
		int no;
		p = in.nextInt();
		n = in.nextInt();
		int[] a = new int[p];
		
		for(i = 0; i < p; i++)	a[i] = 0;
		
		while(n > 0){
			no = in.nextInt();
			if(a[no % p] != 1)	a[no % p] = 1;
				else if(a[no % p] == 1){
					System.out.println(inc);
					flag = 1;
					break;
				}
			inc++;
			n--;
		}
		if(flag == 0)	System.out.println("-1");
	}
}