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 Main {

	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int p=sc.nextInt();
		int n=sc.nextInt();
		int num[]=new int [n];
		boolean[] bucket=new boolean[p];
		int set=-1;
		for(int  i=0;i<n;i++) {
			num[i]=sc.nextInt();
			//System.out.print(num[i]+" ");
		}	
		for(int  i=0;i<n;i++) {
			int culN=num[i]%p;
			//System.out.println("culN="+culN);
			if(bucket[culN]!=true)
				bucket[culN]=true;
			else {
				set=i+1;
				break;
			}	
		}
		System.out.println(set);
		

	}

}
