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
/* @@@@@@@@@@@@@@@@@@@@@@@
 * @@@ Doston Akhmedov @@@
 * @@@@@@@@@@@@@@@@@@@@@@@ 
 */

import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.Locale;
import java.util.StringTokenizer;

public class r_255_a{
	static StringTokenizer st;
	static BufferedReader br;
	static PrintWriter pw;

	public static void main(String[] args) throws IOException {
		Locale.setDefault(Locale.US);
		br = new BufferedReader(new InputStreamReader(System.in));
		pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(
				System.out)));
		int p=nextInt(),n=nextInt(),a[]=new int[n+1];
		boolean used[]=new boolean[10000];
		for (int i = 1; i <=n; i++) {
			a[i]=nextInt();
			a[i]%=p;
		}
		for (int i = 1; i <=n; i++) {			
			if(!used[a[i]])
				used[a[i]]=true;
			else{
				pw.print(i);
				pw.close();
				return;
			}
		}
		pw.print(-1);
		pw.close();
	}

	private static int nextInt() throws IOException {
		return Integer.parseInt(next());
	}

	private static long nextLong() throws IOException {
		return Long.parseLong(next());
	}

	private static double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}

	private static String next() throws IOException {
		while (st == null || !st.hasMoreTokens())
			st = new StringTokenizer(br.readLine());
		return st.nextToken();
	}
}
