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 dzyloveshash
{
    public static void main(String args[])
    {
        Scanner in=new Scanner(System.in);
        int p=in.nextInt();
        int n=in.nextInt();
        int a,i,pos=-1,ans=0;
        int b[]=new int[1000000];
        for(i=0;i<n;i++)
        {
            a=in.nextInt();
            if(ans==0 && b[a%p]==1)
             {
                 ans=1;
                 pos=i;
                }
            b[a%p]=1;
        }
        if(ans==1)
         System.out.println(pos+1);
        else
         System.out.println(-1);
        }
    }
            
