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.io.*;
import java.util.*;

public class p7
{
	public static void main(String[] args)
	{
		FastScanner in = new FastScanner();
		int p = in.nextInt();
		int n = in.nextInt();
		boolean used[] = new boolean[p];
		
		for(int i = 0; i < n; i++)
		{
			int id = in.nextInt() % p;
			if(used[id])
			{
				System.out.println(i+1);
				return;
			}
			
			used[id] = true;
		}
		
		System.out.println(-1);
	}
	
	public static class FastScanner
	{
		BufferedReader br;
		StringTokenizer st;

		public FastScanner(String s)
		{
			try {
				br = new BufferedReader(new FileReader(s));
			} catch (FileNotFoundException e) {
				// TODO Auto-generated catch block
				e.printStackTrace();
			}
		}

		public FastScanner()
		{
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String nextToken()
		{
			while (st == null || !st.hasMoreElements())
			{
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					// TODO Auto-generated catch block
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt()
		{
			return Integer.parseInt(nextToken());
		}

		long nextLong()
		{
			return Long.parseLong(nextToken());
		}

		double nextDouble()
		{
			return Double.parseDouble(nextToken());
		}
		
		String next()
		{
			return nextToken();
		}
	}
}
