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.io.PrintWriter;
import java.util.Scanner;

public class A {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		int p = in.nextInt();
		int n = in.nextInt();
		int[] a = new int[n];
		int[] b = new int[p];
		for (int i = 0; i < p; i++) {
			b[i] = -1;
		}
		for (int i = 0; i < n; i++) {
			a[i] = in.nextInt();
			if (b[a[i] % p] == -1) {
				b[a[i] % p] = 1;
			} else {
				System.out.println(i + 1);
				break;
			}
			if (i == n - 1) {
				System.out.println(-1);
			}
		}
		in.close();
		out.close();
	}

}
