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.Collections;
import java.util.HashSet;
import java.util.Scanner;

import java.util.Vector;

public class Main {

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

	int n=0,p=0;
	p=in.nextInt();
	n=in.nextInt();
	Vector<Integer> nn=new Vector<>();
	for(int i=0;i<n;i++)
	{
		nn.add(in.nextInt());
	}
	
	HashSet<Integer> mm=new HashSet<>();
	
	for(int i=0;i<n;i++)
	{
		if(mm.contains(nn.get(i)%p))
		{
			System.out.println(i+1);
			System.exit(0);
			
		}
		else
		{
			mm.add(nn.get(i)%p);
		}
	}
	System.out.println(-1);
	
	
	
	
	
	
	
	
	
	
	}
	}
