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.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.IOException;
import java.util.StringTokenizer;

public class LovesHash {

	private static class Reader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public  Reader(InputStream input) {
			reader = new BufferedReader(new InputStreamReader(input));
			tokenizer = null;
		}

		public String next() throws IOException{
			while(tokenizer == null || !tokenizer.hasMoreTokens()) {
				tokenizer = new StringTokenizer(reader.readLine());
			}
			return tokenizer.nextToken();
		}

		public int nextInt() throws IOException{
			return Integer.parseInt(next());
		}
	}
	public static void main(String[] args) throws IOException{
		Reader reader = new Reader(System.in);
		int p = reader.nextInt();
		int n = reader.nextInt();
		boolean[] congruence_classes = new boolean[p];
		int ans = -2;
		for(int x = 0; x < n; x++) {
			int mod = reader.nextInt() % p;
			if(congruence_classes[mod] != false) {
				ans = x;
				break;
			}
			else {congruence_classes[mod] = true;}
		}
		System.out.println(ans + 1);
	}
}