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.InputStreamReader;
import java.io.IOException;
import java.lang.Math;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Main {
  public static void main(String[] args) {
    MyScanner sc = new MyScanner();
    int p = sc.nextInt();
    int n = sc.nextInt();
    boolean[] has = new boolean[p];
    for (int i = 1; i <= n; ++i) {
      int cur = sc.nextInt();
      if (has[cur % p]) {
        System.out.println(i);
        return;
      } else has[cur % p] = true;
    }
    System.out.println(-1);
  }
  
  public static class MyScanner {
    BufferedReader br;
    StringTokenizer st;
    public MyScanner() { br = new BufferedReader(new InputStreamReader(System.in)); }
    int nextInt() { return Integer.parseInt(next()); }
    long nextLong() { return Long.parseLong(next()); }
    double nextDouble() { return Double.parseDouble(next()); }
    String next() {
      while (st == null || !st.hasMoreElements()) {
        try { st = new StringTokenizer(br.readLine()); }
        catch (IOException e) { e.printStackTrace(); }
      }
      return st.nextToken();
    }
    String nextLine() {
      String str = "";
      try { str = br.readLine(); }
      catch (IOException e) { e.printStackTrace(); }
      return str;
    }
  }
}