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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;


public class B249 {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String sr = br.readLine();
        StringTokenizer s = new StringTokenizer(sr);
        int p = Integer.parseInt(s.nextToken());
        int nums = Integer.parseInt(s.nextToken());
        int[] hashTab = new int[p];
        int[] num = new int[nums];
        for(int i=0; i<nums; i++) {
            sr = br.readLine();
            num[i] = Integer.parseInt(sr);
        }
        for(int i=0; i< nums; i++) {
            int index = num[i]%p;
            if(hashTab[index] == 0) {
                hashTab[index] = 1;
            } else {
                System.out.println(i+1);
                return;
            }
        }
        System.out.println(-1); 
    }
}
