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
/*
 * To change this template, choose Tools | Templates
 * and open the template in the editor.
 */
//package candidatecode;

//import java.util.*;
import java.io.*;
import java.util.*;

/**
 *
 * @author sandesh
 */
public class Candidatecode {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) throws Exception {
   
    int[] arr=new int[301];
    Arrays.fill(arr, 0);
    Scanner en=new Scanner(System.in);
    int p=en.nextInt();
    int n=en.nextInt();
    int z=0;
    int i=0;
    while(i<n)
    {
        int x=en.nextInt();
        if(arr[x%p]!=0)
        {
            z=i;
            break;
        }
        else
        {
            arr[x%p]++;
        }
        i++;
    }
    if(i==n)
    System.out.println("-1");
    else
    System.out.println(z+1);
    } 
    
}
