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.Scanner;
public class CodeForces {
     
    public static void main(String[] args) {
        try (Scanner scanner = new Scanner(System.in)) {
            int p = scanner.nextInt();
            int n = scanner.nextInt();
            int[] bucket = new int[p];
            int i, x;   
            for(i = 0; i < n; i++){
                x = scanner.nextInt();
                bucket[x % p]++;
                if(bucket[x % p] > 1){
                    System.out.println(i+1);
                    break;
                }
            }
            if(i == n)
                System.out.println("-1");
            scanner.close();
        }
    }
}