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.ArrayList;
import java.util.Scanner;

/**
 * Class: Art<br />
 * Date: 2014/04/06 23:38<br />
 * Description:<br />
 *
 * @author Laiping Zhou(sjtudesigner)
 */
public class Hash {

    public static void main(String[] args)
    {
        Scanner sc = new Scanner(System.in);
        ArrayList<Integer> al = new ArrayList<Integer>();
        int p = sc.nextInt();
        int n = sc.nextInt();

        for (int i = 0;i < n;i++)
        {
            int t = sc.nextInt() % p;
            if (!al.contains(t)) al.add(t);
            else { System.out.println(i + 1); return; }
        }
        System.out.println(-1);
    }

}
