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.math.BigDecimal;
import java.sql.Types;
import java.util.Scanner;

public class dzyloves_hash_447A {
    public static void main(String[] args) {
       
        int bucket=0,num=0;
        Scanner s=new Scanner(System.in);
        bucket=s.nextInt();
        num=s.nextInt();
        int ar[]=new int[num];
        for(int i=0;i<num;i++)
            ar[i]=s.nextInt();
        char ar1[]=new char[bucket];
        int index=0,t=-1;
        for(int i=0;i<num;i++)
        {
            index=ar[i]%bucket;
            if(ar1[index]=='*')
            {t=i;
                break;
            }
            else
            {
                ar1[index]='*';
            }
        }
        if(t!=-1)
        System.out.println((t+1));
        else
            System.out.println("-1");
        
       
    }
    
}
