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
//package CF;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class CFA_FF {
	
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader (System.in));
		PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
		StringTokenizer st = new StringTokenizer(br.readLine());
		boolean[] hash = new boolean[350];
		
		
		int p = Integer.parseInt(st.nextToken());
		int n = Integer.parseInt(st.nextToken());
		int in , sol = -1 ;
		boolean ch = true;
		for (int i = 0; i < n; i++) {
			in = Integer.parseInt(br.readLine());
			if(hash[in%p]&&ch){
				sol = i+1;
				ch = false;
			}
			else
				hash[in%p] = true;
		}
		
		pw.println(sol);
			
		
		pw.close();
		br.close();
	}

}
