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.io.*;
public class CF0447A
{
 public static void main(String[] helloWorld) throws IOException
 {
  BufferedReader sIn = new BufferedReader(new InputStreamReader(System.in));
  String iS = sIn.readLine();
  int space = iS.indexOf(32);
  int p = Integer.parseInt(iS.substring(0, space));
  int n = Integer.parseInt(iS.substring(space + 1));
  boolean[] taken = new boolean[p];
  for (int i = 1; i <= n; i ++)
  {
   int c = Integer.parseInt(sIn.readLine());
   if (taken[c % p])
   {
    System.out.println(i);
    return;
   }
    else
     taken[c % p] = true;
  }
  System.out.println("-1");
 }
}
