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.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.*;

public class B implements Runnable {
    int n, k;

    private void solve() throws IOException {
        n = nextInt();
        k = nextInt();
        if (n == k) {
            pl(-1);
            return;
        }
        if (n == k + 1) {
            for (int i = 1; i <= n; i++) {
                p(i + " ");
            }
            pl();
            return;
        }
        if (k == 0) {
            for (int i = 2; i <= n; i++)
                p(i + " ");
            pl(1);
            return;
        }
        StringBuilder sb = new StringBuilder();
        int cnt = 0, i = 1;
        for (i = 2; i <= n; i++) {
            sb.append(i + " ");
            ++cnt;
            if (cnt == k) {
                break;
            }
        }
        for (int j = i + 1; j < n; j++) {
            sb.append((j + 1) + " ");
        }
        p((i + 1) + " ");
        sb.append(1);
        pl(sb);
    }

    private int gcd(int a, int b) {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }

    public static void main(String[] args) {
        new B().run();
    }

    BufferedReader reader;
    StringTokenizer tokenizer;
    PrintWriter writer;

    public void run() {
        try {
            reader = new BufferedReader(new BufferedReader(
                    new InputStreamReader(System.in)));
            writer = new PrintWriter(System.out);
            tokenizer = null;
            solve();
            reader.close();
            writer.close();
        } catch (Exception e) {
            e.printStackTrace();
            System.exit(1);
        }
    }

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

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

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

    BigInteger nextBigInteger() throws IOException {
        return new BigInteger(nextToken());
    }

    String nextToken() throws IOException {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            tokenizer = new StringTokenizer(reader.readLine());
        }
        return tokenizer.nextToken();
    }

    void p(Object... objects) {
        for (int i = 0; i < objects.length; i++) {
            if (i != 0)
                writer.print(' ');
            writer.print(objects[i]);
        }
    }

    void pl(Object... objects) {
        p(objects);
        writer.println();
    }

    int cc;

    void pf() {
        writer.printf("Case #%d: ", ++cc);
    }

}