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.*;
public class a447
{
	public static void main(String ar[])
	{
		Scanner ob=new Scanner(System.in);
		int p=ob.nextInt();
		int n=ob.nextInt();
		int ans=0;
		HashSet<Integer> myset=new HashSet<Integer>();
		for(int i=0;i<n;i++)
		{
			int x=ob.nextInt();
			x=x%p;
			if(myset.contains(x) && ans==0)
			{
				ans=i+1;
			}
			else
			{
				myset.add(x);
			}
		}
		if(ans>0)
			System.out.println(ans);
		else
		System.out.println(-1);
	}
}
