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.Scanner;


public class NewClass {
    static int answer = -1;
    public static void main (String[]args){
        Scanner in = new Scanner(System.in);
        String k = in.nextLine();
        String[] k1 = k.split(" ");
        int size = Integer.parseInt(k1[0]);
        int zanyato = Integer.parseInt(k1[1]);
        //System.out.println(size+ "  " +zanyato);
        int[] a1 = new int[size];
        //ArrayList<Integer> a = new ArrayList<Integer>(size);
        for(int i=0;i<zanyato;i++){
            int index = in.nextInt()%size;
            if(a1[index]==0)a1[index]++;
            else {answer=i+1;break;}
        //if(a.contains(index))////    print(a1);
        }
        System.out.println(answer);
    }
    static void print(int []a){
        for(int i=0;i<a.length;i++){
            System.out.print(a[i]+" ");
        }
        System.out.println("");
    }
}
