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.ArrayList;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        
        Scanner input=new Scanner(System.in);
        int p=input.nextInt();
        int n=input.nextInt();
        int x;
        int[] ans=new int[n];
        for(int i=0;i<n;i++)
        {
            x=input.nextInt();
            ans[i]=x%p;
        }
        int answ=love(n,ans);
        if(answ>0)
         {
           System.out.println(answ+1);
         }
        else
        {
            System.out.println(-1);
        }
    }
    
    
    public static int love(int n,int[] ans){
        ArrayList<Integer> lo=new ArrayList<Integer>();
           for(int j=0;j<n;j++)
            {
              for(int k=j+1;k<n;k++)
                {
                   if(ans[j]==ans[k])
                   {
                       lo.add(k);
                   }
                }
            }
          return one(lo);   
     }
    
    public static int one(ArrayList lo)
    {
        int min=10000;
        for(int l=0;l<lo.size();l++)
        {
            if((int)(lo.get(l))<min)
            {
                min=(int)lo.get(l);
            }
        }
        if(min<10000)
        {
        return min;
        }
        else
        {
            return 0;
        }
    }

}
