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.*;
import java.util.*;

public class Main{
  final boolean isFileIO = false; 
  BufferedReader in;
  PrintWriter out;
  StringTokenizer st = new StringTokenizer("");
  String delim = " ";

  public static void main(String[] args) throws IOException {
    Main m = new Main();
    m.initIO();
    m.solve();
    m.in.close();
    m.out.close();
  }

  public void initIO() throws IOException {
    if(!isFileIO) {
      in = new BufferedReader(new InputStreamReader(System.in));
      out = new PrintWriter(System.out);
    } else {
      in = new BufferedReader(new FileReader("A-large-practice.in"));
      out = new PrintWriter("output.txt"); 
    }
  }
  
  String nextToken() throws IOException {
    if(!st.hasMoreTokens()) {
      st = new StringTokenizer(in.readLine());
    }
    return st.nextToken(delim);
  }
 
  String readLine() throws IOException {
    return in.readLine();
  }

  int nextInt() throws IOException {
   return Integer.parseInt(nextToken());
  }

  long nextLong() throws IOException {
    return Long.parseLong(nextToken());
  }

  double nextDouble() throws IOException {
    return Double.parseDouble(nextToken());
  }

  void solve() throws IOException {
    int p, n;
    p = nextInt(); n = nextInt();
    boolean[] u = new boolean[p];
    for(int i = 0; i < n; i++) {
      int x = nextInt();
      if(u[x % p]) {
        out.println(i + 1);
        return;
      } else {
        u[x % p] = true;
      }
    }
    out.println(-1);
  }
  
}
