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.*;
import java.util.*;


public class Task1 {

	public static void main(String[] args) throws Exception {
		solve();
	}

	private static void solve() throws Exception {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		//PrintWriter out = new PrintWriter(new );
		StringTokenizer tok = new StringTokenizer(in.readLine());
		int p = Integer.parseInt(tok.nextToken());
		int n = Integer.parseInt(tok.nextToken());
		int mas[] = new int[p];
		Arrays.fill(mas, -1);
		int res = -1;
		for (int i = 0; i < n; i++) {
			int tmp = Integer.parseInt(in.readLine());
			if (mas[tmp % p] == -1) mas[tmp % p] = 1; 
				else {
					res = i+1;
					break;
				}
		}
		System.out.println(res);
	}

}
