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 Main {
    
    public static void main(String[] args) {
        
        Scanner input = new Scanner (System.in);
        
        int p = input.nextInt();
        int n = input.nextInt();
        
        int [] array = new int [p];
        
        Arrays.fill(array , -1);
        
        boolean hash = true;
        
        for(int i = 0 ; i < n ; i++)
        {
            int x = input.nextInt();
            
            int rem = x % p;
            
            if(array[rem] == -1)
                array[rem] = x;
            else
            {
                System.out.println(i+1);
                hash = false;
                break;
            }
        }
        
        if(hash)
            System.out.println("-1");
    }
    
}
