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.Collections;
import java.util.Scanner;

public class Main {


    public static void main(String[] args) {

   Scanner scr = new Scanner(System.in);
        int p =scr.nextInt();
        int n = scr.nextInt();
        boolean flag = true;
        ArrayList<Integer> a = new ArrayList<Integer>(n);
      ArrayList<Integer> hesArray =new ArrayList<Integer>(n);
        for (int i=0;i<n;i++) {

            a.add(scr.nextInt());

        }

 int temp=0;
        for (int i=0;i<n;i++){

            temp=a.get(i) % p;

            if (i!=0){
            if(hesArray.contains(temp)){
                System.out.println(i+1);
                flag=false;
                break;
            }
        }

            hesArray.add(temp);


        }


  if (flag) System.out.println("-1");

    }

}