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
/*
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import java.util.Scanner;

/**
 *
 * @author mostafa.ayman
 */
public class Codeforces {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        // TODO code application logic here
        Scanner in  = new Scanner(System.in);
        
        int p = in.nextInt();
        int n = in.nextInt();
        
        int buckets[] = new int[p];
        for(int i=0;i<n;i++)
        {
            int num = in.nextInt()%p;
           if(buckets[num]>0)
           {
               System.out.println(i+1);
               return;
           }
           buckets[num]++;
        }
        
        System.out.println(-1);
        
    }
    
}
