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.*;
import java.io.*;
public class CF447A {
	public static void main(String[] args) throws IOException {
		in.init(System.in);
		PrintWriter out = new PrintWriter(System.out);

		int p = in.nextInt();
		int n = in.nextInt();
		boolean[] used = new boolean[p];
		int ans = -2;
		for(int i = 0; i < n; i ++)
		{
			int a = in.nextInt()%p;
			if(used[a])
			{
				ans = i;
				break;
			}
			used[a] = true;
		}
		out.println(ans + 1);
		out.close();
	}

	public static class in {
		static BufferedReader reader;
		static StringTokenizer tokenizer;

		static void init(InputStream in) {
			reader = new BufferedReader(new InputStreamReader(in));
			tokenizer = new StringTokenizer("");
		}

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

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

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

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