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.*;
public class A447
{
    public static void main(String args[])
    {
        Scanner in = new Scanner(System.in);
        int p = in.nextInt();
        int n = in.nextInt();
        int a[] = new int [n];
        int h[] = new int [p];
        int c=0;
        int i;
        int mod = 0;
        for( i = 0 ; i < n ; i++)
        {
            a[i] = in.nextInt();
            mod = a[i] % p;
            if( h[mod] == 0)
            {
                h[mod] = -2 ;
            }
            else
            {
                System.out.println(i+1);
                c=1;
                break;
             }
        }
        if( c==0)
        {
        System.out.println("-1");
    }
    }
}
        
        