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.Scanner;

/**
 *
 * @author Admin
 */
public class DZY {

    public static void main(String[] args) {
        int flag = 0;
        boolean ok = false;
        int k;
        char[] m = new char[305];
        Scanner in = new Scanner(System.in);
        int p = in.nextInt();
        int n = in.nextInt();
        
        for (int i = 0; i < p; i++) {
            m[i] = 0;
        }

        for (int i = 0; i < n; i++) {
            int x = in.nextInt();
            k = x % p;
            if (m[k] == 0) {
                m[k] = 1;
            } 
            else if (!ok) {
                flag = i + 1;
                ok = true;
            }
        }
        if (ok) {
            System.out.println(flag);
        } else {
            System.out.println(-1);
        }
    }
}