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 probA {
public static void main (String[]args)throws IOException{
	BufferedReader x = new BufferedReader(new InputStreamReader(System.in));
	StringTokenizer st=new StringTokenizer(x.readLine());
	int p=Integer.parseInt(st.nextToken());
	int n=Integer.parseInt(st.nextToken());
	HashSet <Integer>possible=new HashSet<Integer>();
	for (int i=0; i<n; i++){
		int current=Integer.parseInt(x.readLine())%p;
		if (possible.contains(current)){
			System.out.println(i+1);
			System.exit(0);
		}
		else{
			possible.add(current);
		}
	}
	System.out.println("-1");
	System.exit(0);
}
}