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.*;
import java.io.*;
public class DZY_Loves_Hash
{
	public static void main(String args[]) throws Exception
	{
		BufferedReader f=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer(f.readLine());
		int mod=Integer.parseInt(st.nextToken());
		int runs=Integer.parseInt(st.nextToken());
		boolean[] arr=new boolean[mod];
		boolean flag=true;
		for(int x=0;x<runs;x++)
		{
			int temp=Integer.parseInt(f.readLine());
			if(arr[temp%mod])
			{
				flag=false;
				System.out.println(x+1);
				break;
			}
			arr[temp%mod]=true;
		}
		System.out.println(flag?"-1":"");
	}
}