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 Main {

  public static void main (String[] args) {

    Scanner sc = new Scanner(System.in);
    int p = sc.nextInt();
    int n = sc.nextInt();
    
    int answer = -1;

    int[] h = new int[p];

    Arrays.fill(h, -1);

    for (int i = 0; i < n; i++) {
      int x = sc.nextInt();
      if (h[x % p] == x % p) {
        if (answer == -1) {
          answer = i + 1;
        }
      } else {
        h[x % p] = x % p;
      }
    }
    sc.close();

    System.out.println(answer);

  }

}