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.InputStreamReader;
import java.util.HashSet;
import java.util.Set;

public class Main {

    /**
     * @param args
     */
    public static void main(String[] args) {
        try {
            BufferedReader br = new BufferedReader(new InputStreamReader(
                    System.in));
            String[] strArray = br.readLine().split(" ");
            int p = Integer.parseInt(strArray[0]);
            int n = Integer.parseInt(strArray[1]);
            Set<Integer> set = new HashSet<Integer>();
            int rtn = -1;
            for (int i = 0; i < n; i++) {
                int x = Integer.parseInt(br.readLine());
                int h = x % p;
                if(set.contains(h)){
                    rtn = i + 1;
                    break;
                }else{
                    set.add(h);
                }
            }

             System.out.println(rtn);
        } catch (Exception e) {
            System.err.println("Error:" + e.getMessage());
        }
    }
}
