Levko loves permutations very much. A permutation of length n is a sequence of distinct positive integers, each is at most n.

Let’s assume that value gcd(a, b) shows the greatest common divisor of numbers a and b. Levko assumes that element pi of permutation p1, p2, ... , pn is good if gcd(i, pi) > 1. Levko considers a permutation beautiful, if it has exactly k good elements. Unfortunately, he doesn’t know any beautiful permutation. Your task is to help him to find at least one of them.

Input

The single line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ n).

Output

In a single line print either any beautiful permutation or -1, if such permutation doesn’t exist.

If there are multiple suitable permutations, you are allowed to print any of them.

Examples

Input

4 2


Output

2 4 3 1

Input

1 1


Output

-1

Note

In the first sample elements 4 and 3 are good because gcd(2, 4) = 2 > 1 and gcd(3, 3) = 3 > 1. Elements 2 and 1 are not good because gcd(1, 2) = 1 and gcd(4, 1) = 1. As there are exactly 2 good elements, the permutation is beautiful.

The second sample has no beautiful permutations.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class B {

  public static void main(String[] args) {
    solve(new InputReader(System.in));
  }

  static void solve(InputReader in) {
    long n = in.ni();
    long k = in.ni();

    if (n == k) {
      System.out.println(-1);
      return;
    }

    StringBuilder builder = new StringBuilder();

    for (long i = 1; i < n - k; i++) {
      builder.append(i + 1 + " ");
    }
    builder.append(1 + " ");
    for (long i = n - k + 1; i <= n; i++) {
      builder.append(i + " ");
    }

    System.out.println(builder);
  }

  static class InputReader {
    BufferedReader br;
    StringTokenizer st;

    InputReader(InputStream stream) {
      br = new BufferedReader(new InputStreamReader(stream));
    }

    String next() {
      while (st == null || !st.hasMoreElements()) {
        try {
          st = new StringTokenizer(br.readLine());
        } catch (IOException e) {
          e.printStackTrace();
        }
      }
      return st.nextToken();
    }

    int ni() {
      return Integer.parseInt(next());
    }

    long nl() {
      return Long.parseLong(next());
    }

    int[] na(int n) {
      int[] arr = new int[n];
      for (int i = 0; i < n; i++) {
        arr[i] = ni();
      }
      return arr;
    }
  }
}
