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.Scanner;

public class problem60 {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
Scanner sc = new Scanner(System.in);
//System.out.println("sjbc");
int  p  = sc.nextInt();
int n = sc.nextInt();
int ar[] = new int[p];

//HashSet<Integer> hs= new HashSet<>();
int c=0;
int q=0;
for(int i=0;i<n;i++)
{
int x = sc.nextInt();
int y = x%p;
if(ar[y]!=1)
{
ar[y] = 1;
c++;
	}
else
{
	System.out.println(c+1);
	q=1;
	break;
}}
if(q!=1)
{
	System.out.println("-1");
}



	

	}}