Problem: 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.

Solution (in JAVA):

Solution:
import java.io.IOException;
import java.io.InputStream;

public class Main {
    public static void main(String[] args) throws IOException {
        FastScanner s = new FastScanner(System.in);
        int n = s.nextInt();
        int k = s.nextInt();
        if (k == n) {
            System.out.print(-1);
        } else {
            int t = n - k;
            if (t == 1) {
                for (int i = t; i <= n; i++) {
                    System.out.print(i + " ");
                }
            } else {
                System.out.print(n + " ");
                for (int i = 2; i < 2 + k; i++) {
                    System.out.print(i + " ");
                }
                System.out.print(1 + " ");
                for (int i = 2 + k; i <= n - 1; i++) {
                    System.out.print(i + " ");
                }

            }


        }

    }
}

class FastScanner {

    InputStream is;
    byte buff[] = new byte[1024];
    int currentChar = -1;
    int buffChars = 0;

    public FastScanner(InputStream inputStream) {
        is = inputStream;
    }

    public boolean hasNext() throws IOException {
        return buffChars >= 0;
    }

    public int nextChar() throws IOException {
        //if we already have that next char read, just return else input
        if (currentChar == -1 || currentChar >= buffChars) {
            currentChar = 0;
            buffChars = is.read(buff);
        }

        if (buffChars <= 0) {
            return -1;
        }

        return (char) buff[currentChar++];
    }

    public String nextString() throws IOException {
        StringBuilder bldr = new StringBuilder();
        int ch;
        while (isSpace(ch = nextChar())) ;

        do {
            bldr.append((char) ch);
        } while (!isSpace(ch = nextChar()));

        return bldr.toString();
    }

    public int nextInt() throws IOException {
        //considering ASCII files--> 8 bit chars,     unicode files has 16 bit chars (byte1 then byte2)
        int result = 0;
        int sign = 1;
        int ch;
        while (isSpace(ch = nextChar())) ;

        if (ch == '-') {
            sign = -1;
            ch = nextChar();
        }

        do {
            if (ch < '0' || ch > '9') {
                throw new NumberFormatException("Found '" + ch + "' while parsing for int.");
            }

            result *= 10;
            result += ch - '0';
        } while (!isSpace(ch = nextChar()));

        return sign * result;
    }

    public long nextLong() throws IOException {
        //considering ASCII files--> 8 bit chars,     unicode files has 16 bit chars (byte1 then byte2)
        long result = 0;
        int sign = 1;
        int ch;
        while (isSpace(ch = nextChar())) ;

        if (ch == '-') {
            sign = -1;
            ch = nextChar();
        }

        do {
            if (ch < '0' || ch > '9') {
                throw new NumberFormatException("Found '" + ch + "' while parsing for int.");
            }

            result *= 10;
            result += ch - '0';
        } while (!isSpace(ch = nextChar()));

        return sign * result;
    }

    private boolean isSpace(int ch) {
        return ch == '\n' || ch == ' ' || ch == '\t' || ch == '\r' || ch == -1;
    }

    public void close() throws IOException {
        is.close();
    }
}