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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;



public class DZYLovesHash {
    public static void main(String[] args) throws NumberFormatException, IOException {
          BufferedReader br = new BufferedReader( new InputStreamReader( System.in ) );
            
            String s = br.readLine();
            String [] s1 =s.split(" ");
            
            int [] hash = new int [Integer.parseInt(s1[0])];
            for(int i=0; i<hash.length; i++){
                hash[i]=-1;
            }
            //ArrayList <Integer> hash = new ArrayList<Integer>(); 
            int v =0;
            int c =1;
            
            while(br.ready()) {
                int x = Integer.parseInt(br.readLine());
                v=x%Integer.parseInt(s1[0]);
                
                if(hash[v] != -1) {
                    System.out.println(c);
                    return;
                }
                else {
                    hash[v]=v;
                }
                c++;
            }
            
            System.out.println(-1);
            
    }
}
