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
/*
 * Code Author: Akshay Miterani
 * DA-IICT
 */
import java.io.*;
import java.util.*;

public class MainB {

	
	public static void main(String args[]){
		InputReader in = new InputReader(System.in);
		OutputStream outputStream = System.out;
		PrintWriter out = new PrintWriter(outputStream);
		//----------My Code Starts Here----------
		int p=in.nextInt();
		int n=in.nextInt();
		int a[]=new int[n];
		int h[]=new int[p];
		for(int i=0;i<n;i++){
			a[i]=in.nextInt();
			int hash=a[i]%p;
			if(h[hash]!=0){
				System.out.println(i+1);
				return;
			}
			h[hash]=1;
		}
		System.out.println("-1");
		out.close();
		//---------------The End------------------
	}
	
	static class InputReader {
	    public BufferedReader reader;
	    public StringTokenizer tokenizer;

	    public InputReader(InputStream inputstream) {
	      reader = new BufferedReader(new InputStreamReader(inputstream));
	      tokenizer = null;
	    }
	    
	    public String nextLine(){
	    	String fullLine=null;
	    	while (tokenizer == null || !tokenizer.hasMoreTokens()) {
	            try {
	              fullLine=reader.readLine();
	            } catch (IOException e) {
	              throw new RuntimeException(e);
	            }
	            return fullLine;
	          }
	          return fullLine;
	    }
		public String next() {
	      while (tokenizer == null || !tokenizer.hasMoreTokens()) {
	        try {
	          tokenizer = new StringTokenizer(reader.readLine());
	        } catch (IOException e) {
	          throw new RuntimeException(e);
	        }
	      }
	      return tokenizer.nextToken();
	    }
		public long nextLong() {
		      return Long.parseLong(next());
		    }
	    public int nextInt() {
	      return Integer.parseInt(next());
	    }
	  }
}