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