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 A {

	private static Scanner in;
	
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		in = new Scanner(System.in);
		long p = in.nextLong();
		long n = in.nextLong();
		long store[]=new long[(int) p];
		Arrays.fill(store, -1);
		boolean conflict=false;
		long loc = 0;
		
		for(int i=0;i<n;i++)
		{
			long x = in.nextLong();
			if(store[(int) (x%p)]==-1)
				store[(int) (x%p)]=(int) x;
			else
			{
				conflict=true;
				loc=i;
				break;
			}	
		}
		if(conflict==true)
			System.out.println(loc+1);
		else
			System.out.println("-1");
	}

}
