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.Scanner;


public class A {
public static void main(String[] args) {
	Scanner in = new Scanner(System.in);
	int p=in.nextInt();
	int n=in.nextInt();
	boolean[] ar=new boolean [p];
	int ans=0;
	boolean c=false;
	for (int i = 0; i < n; i++) {
		int x=in.nextInt();
		
		if(!c&&ar[x%p]){
			ans=i;
			c=true;
//			break;
		}
		ar[x%p]=true;
		
	}
	if(!c)
	System.out.println(-1);
	else
		System.out.println(ans+1);
	
}
}
