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.HashSet;
import java.util.Scanner;
import java.util.Set;

public class Main{
    public static void main(String[] args){
        Scanner scanner = new Scanner(System.in);
        //LinkedHashMap<Integer, Integer> map = new LinkedHashMap<>();
        Set<Integer> set = new HashSet<>();
        int p = scanner.nextInt();
        int n = scanner.nextInt();
        int count = 0;
        boolean a=false;
        boolean check =true;
        int[] arr= new int[n];
        for (int i=0;i<n;i++) {
            arr[i] =scanner.nextInt();
            if (!set.add(arr[i] % p) && !a){
                check = false;
                count=i+1;
                a=true;

            }
        }

        if(check){
            System.out.println(-1);
        }else {
            System.out.println(count);
        }


    }
}