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.Arrays;
import java.util.Scanner;

/**
 *
 * @author ganeshpachpind
 */
public class CodeForce447A {
    public static void main(String args[])
    {
        CodeForce447A process=new CodeForce447A();
        process.doProcess();
    }
    void doProcess()
    {
        Scanner sc=new Scanner(System.in);
        int p=sc.nextInt();
        int n=sc.nextInt();
        int a[]=new int[p];
        Arrays.fill(a,-1);
        System.out.println(check(p, a, sc,n));

    }
    int check(int p,int a[],Scanner sc,int n)
    {
        for(int i=0;i<n;i++)
        {
            int mod=(sc.nextInt())%p;
            if(a[mod]==-1)
            {
                a[mod]=mod;
            }
            else
            {                
                return i+1;                
            }            
        }
        return -1;
    }
    
}
