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.*;


public class round447problema {
	static Scanner in = new Scanner(System.in);

	public static void main(String[] args) {
	 int bucket=in.nextInt();
	 int count=in.nextInt();
	 HashSet<Integer> set=new HashSet<Integer>();
	 for(int i=0;i<count;i++){
		 int num=in.nextInt();
		 int rem=num%bucket;
		 if(set.contains(rem)){
			 System.out.print(i+1);
			 return;
		 }
		 else{
			 set.add(rem);
		 }
	 }
	
	System.out.println(-1);
	}
}
