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.*;
import java.math.*;

public class Exercise1 implements Runnable {

    static BufferedReader in;
    static PrintWriter out;
    static StringTokenizer st;

    private void solve() throws IOException {
        int divisor = nextInt();
        int count = nextInt();

        HashSet<Integer> hashes = new HashSet<>();
        Integer collision = null;

        for (int i = 0; i < count && collision == null; i++) {
            int mod = nextInt() % divisor;

            if (!hashes.contains(mod)) {
                hashes.add(mod);
            } else {
                collision = i;
            }
        }

        if (collision != null) {
            out.print(collision + 1);
        } else {
            out.print(-1);
        }
    }

    public static void main(String[] args) {
        new Exercise1().run();
    }

    @Override
    public void run() {
        try {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);

            solve();

            out.close();
        } catch (IOException e) {
            e.printStackTrace(System.err);
            System.exit(1);
        }
    }

    private String nextToken() throws IOException {
        while (st == null || !st.hasMoreTokens()) {
            String line = in.readLine();

            if (line == null) {
                return null;
            }

            st = new StringTokenizer(line);
        }

        return st.nextToken();
    }

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

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

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