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.Arrays;
import java.util.Scanner;

public class DZYLovesHash {
    public static void main(String asd[])throws Exception
    {
        Scanner in=new Scanner(System.in);
        int p=in.nextInt();
        int n=in.nextInt();
        int a[]=new int[p];
        int b[]=new int[n];
        int c=-1;
        Arrays.fill(a,0);
        for(int i=0;i<n;i++)
        {
            b[i]=in.nextInt();
        }
        for(int i=0;i<n;i++)
        {
            int x=b[i]%p;
           if(a[x]==0)
                a[x]=1;
            else
            {
                c=i+1;
                break;
            }
        }
        if(c>0)
            System.out.println(c);
        else
            System.out.println(-1);
    }
}
