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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashSet;
import java.util.Scanner;
import java.util.StringTokenizer;
public class Main {

    public static void main(String[] args) throws IOException {
        // TODO Auto-generated method stub

        int p,n;
        Scanner sc=new Scanner(System.in);
        HashSet<Integer> hs=new HashSet<Integer>();
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
    
         p=sc.nextInt();
         n=sc.nextInt();
    
        int arr[]=new int[n];
        int brr[]=new int[n];
        for(int i=0;i<n;i++)
            arr[i]= Integer.parseInt(sc.next());
        for(int i=0;i<n;i++)
        {
            brr[i]=arr[i]%p;
        }
        int count=1;
        for(int i=0;i<n;i++)
        {
            if(hs.add(brr[i])==true)
            {
                count++;
            }
            else break;
        }
        //System.out.println(brr.length);
        if(count==(brr.length+1))
            System.out.println(-1);
        else System.out.println(count);
    }

}
