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.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;



public class CodeForces241 {
    
    
    
    public static void main(String[] args) throws IOException {
        
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bw = new BufferedWriter(new PrintWriter(System.out));
        
        String[] ln1 = br.readLine().split(" ");
        int p = Integer.parseInt(ln1[0]);
        int n = Integer.parseInt(ln1[1]);
        
        int[] b = new int[p];
        int[] nu = new int[n];
        
        for(int i=0;i<n;i++){
            nu[i]=Integer.parseInt(br.readLine());
        }
        int val=-1;
        for(int i=0;i<n;i++){
            int elem = nu[i];
            int pos = elem%p;
            if(b[pos]==1){
                val=i+1;
                break;
            }else{
                b[pos]=1;
            }
        }
        
        bw.append(val+"\n");
           bw.close();     
    }
}