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.*;

public class DZYHash447A
{
    public static void main(String[] args) 
    {
        // Set up scanner
        Scanner sc = new Scanner(System.in); 
        // System.out.println("Input p");
        int p = sc.nextInt();
        // System.out.println("Input n");
        int n = sc.nextInt();
        
        boolean[] a = new boolean[p];
        
        for (int i=1; i<=n; i++)
        {
            // System.out.println("Input next value");
            int x = sc.nextInt();
            int hashed = x % p;
            if (a[hashed] == true)  // Already placed?
            {
                System.out.println(i);
                return;
            }
            else
            {
                a[hashed] = true;
            }
        }
        System.out.println(-1);
    }
}
        
            
            
        
        