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 A {
	FastScanner in;
	PrintWriter out;

	void solve() {
		boolean used[] = new boolean[333];
		int p = in.nextInt(), n = in.nextInt();
		for (int i = 0; i < n; i++) {
			int a = in.nextInt();
			if (used[a % p]) {
				out.print(i + 1);
				return;
			}
			used[a % p] = true;
		}
		out.print(-1);
		
	}

	public static void main(String args[]) {
		new A().runIO();
	}

	void run() {
		try {
			in = new FastScanner(new File("Timus.in"));
			out = new PrintWriter(new File("Timus.out"));

			solve();

			out.close();
		} catch (FileNotFoundException e) {
			e.printStackTrace();
		}
	}

	void runIO() {
		in = new FastScanner(System.in);
		out = new PrintWriter(System.out);

		solve();

		out.close();
	}

	class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		FastScanner(File f) {
			try {
				br = new BufferedReader(new FileReader(f));
			} catch (FileNotFoundException e) {
				e.printStackTrace();
			}
		}

		FastScanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		String next() {
			while (st == null || !st.hasMoreTokens()) {
				String s = null;
				try {
					s = br.readLine();
				} catch (IOException e) {
					e.printStackTrace();
				}
				if (s == null)
					return null;
				st = new StringTokenizer(s);
			}
			return st.nextToken();
		}

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

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

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