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
/**
 * Created by Phillip on 7/13/2014.
 */
import java.io.*;
public class Main {
    public static void main(String[] args) throws IOException{
        BufferedReader stdin = new BufferedReader(new InputStreamReader(System.in));
        String[] input = stdin.readLine().split(" ");
        int p = Integer.parseInt(input[0]);
        int n = Integer.parseInt(input[1]);
        boolean[] modulo = new boolean[p];
        int[] numbers = new int[n];
        for(int i=0;i<n;i++){
            numbers[i] = Integer.parseInt(stdin.readLine());
        }
        loop(p,n,modulo,numbers);
    }
    public static void loop (int p,int n,boolean[] modulo,int[] numbers) throws IOException{
        for(int i=0;i<n;i++){
            if(!modulo[numbers[i]%p]){
                modulo[numbers[i]%p] = true;
            }
            else{
                System.out.println(i+1);
                return;
            }
        }
        System.out.println(-1);
    }
}
