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.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.Scanner;


public class mg8 {
	public static void main(String[] args){
		Scanner in = new Scanner(System.in);
		int n =0;
        int p =0;
       String pn = in.nextLine();
       String[] arraypn = pn.split(" ");
       p = Integer.parseInt(arraypn[0]);
       n= Integer.parseInt(arraypn[1]);
       HashMap<Integer,Integer> s = new HashMap<Integer,Integer>();
       for(int i =0;i<p;i++){
    	   s.put(i, -1);
       }
       for(int i =0;i<n;i++){
    	   int x = Integer.parseInt(in.nextLine());
    	   if(s.get(x%p)==-1){
    		   s.put(x%p,x);
    	   }
    	   else {System.out.println(i+1);
    	   return;} 
       }
       System.out.println(-1);
       return;
	}

}
