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.io.OutputStreamWriter;
import java.io.PrintWriter;


public class DZYLovesHash {
    public static void main(String[] args) throws IOException {
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        String s[]=in.readLine().split(" ");
        int p=Integer.parseInt(s[0]);
        int n=Integer.parseInt(s[1]);
        int res=-1;
        boolean num[]=new boolean[p];
        for (int i = 0; i < n; i++) {
            int k=Integer.parseInt(in.readLine());
            int hash=k%p;
            if(num[hash]&&res==-1){
                res=i+1;
            }else{
                num[hash]=true;
            }
            
        }
        out.println(res);
        out.close();
    }

}
