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.*;
import java.util.*;


public class Main {
   
    public static void main(String[] args)throws IOException {
        
        int n,p;
        long x;
        BufferedReader in=new BufferedReader(new InputStreamReader(System.in));
        String s;
        StringTokenizer token;
            token=new StringTokenizer(in.readLine()," ");
            p=Integer.parseInt(token.nextToken());
            n=Integer.parseInt(token.nextToken());
            int V[]=new int[p];
        //  Arrays.fill(V,-1);
            boolean conflic=false;
            int pos=-1;
            x=Integer.parseInt(in.readLine());
            V[(int)(x%p)]=1;
            for(int i=1;i<n;i++){
                x=Integer.parseInt(in.readLine());
                
                if(V[(int)(x%p)]!=0){
                    conflic=true;
                    pos=i+1;
                    break;
                }
                //System.out.println(x%p);
                V[(int)(x%p)]=1;                    
            }

        System.out.println(pos);
            
    }
}