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.HashSet;
import java.util.Hashtable;
import java.util.Scanner;
import java.util.Vector;

public class Main{

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        Vector<Integer >v = new Vector<Integer>();
        Vector<Integer >v2 = new Vector<Integer>();
        int p =sc.nextInt();
        int n = sc.nextInt();
        boolean f= false;
        int index =0;
        for(int i = 0,j=0 ;j<n ;++i,j++)
        {
            int num = sc.nextInt();
            
                if(v2.contains(num%p))
                {
                index =j+1;
                f=true;
                break;
                }
            
        v2.add(num%p);  
            
        }
        if(f)System.out.println(index);
        else
        System.out.println(-1);
        
        
        
    }

}
