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 Problem {
        public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int p = input.nextInt();
        int n = input.nextInt();
        long [] array = new long[n];
        for ( int i = 0 ; i < n ; i++ )
            array[i] = input.nextLong();
        int [] times = new int[p];
        long temp ;
        long counter = 1;
        for ( int i = 0 ; i < n ; i++ ){
            if ( array[i] != 0 )
            {
                temp = array[i]%p;
                times[(int)temp]++;
                if ( times[(int)temp] > 1 )
                {
                    System.out.println(counter);
                    return;
                }
            }else{
                times[0]++;
                if ( times[0] > 1 )
                {
                    System.out.println(counter);
                    return;
                }
            }
            counter++;
        }
        System.out.println(-1);
    }
}