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 DZYSucksAtExplaining {

	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int p = scan.nextInt();
		int n = scan.nextInt();
		scan.nextLine();
		int[] A = new int[n];
		for(int i = 0; i < n; i++) {
			A[i] = scan.nextInt() % p;
			scan.nextLine();
		}
		ArrayList<Integer> pool = new ArrayList<Integer>();
		for(int i = 0; i < n; i++) {
			if(pool.contains(A[i])) {
				System.out.println(i+1);
				return;
			}
			pool.add(A[i]);
		}
		System.out.println(-1);
	}

}
