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.OutputStream;
import java.io.PrintWriter;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

/**
 * Built using CHelper plug-in Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
	OutputStream outputStream = System.out;
	Scanner in = new Scanner(System.in);
	PrintWriter out = new PrintWriter(outputStream);
	Problem solver = new Problem();
	solver.solve(in, out);
	out.close();
    }
}

class Problem {
    
    Map<Integer, Integer> map = new HashMap<Integer, Integer>();

    public void solve(Scanner in, PrintWriter out) {
	Integer p = in.nextInt();
	Integer n = in.nextInt();
	boolean conflict = false;
	
	for (int i = 1; i <= n ; i++) {
	    Integer toInsert = in.nextInt();
	    int hx = toInsert % p;
	    
	    if (map.containsKey(hx)) {
		out.println(i);
		conflict = true;
		break;
	    } else {
		map.put(hx, toInsert);
	    }
	}
	
	if (!conflict) {
	    out.println("-1");
	}
    }

}