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.*;
import java.math.BigInteger;
import java.util.*;
import java.util.Collections.*;

import static java.util.Collections.*;
import static java.lang.Math.*;
import static java.util.Arrays.*;
import static java.math.BigInteger.*;
import static java.lang.Character.*;

public class Main {

    BufferedReader br;
    StringTokenizer st;
    PrintWriter out;

    void run(){
        Locale.setDefault(Locale.US);
        boolean oj = System.getProperty("ONLINE_JUDGE")!=null;
//        boolean oj = true;
        String fileName = oj? "curiosity" : "dwarf";
        try {
            if( oj ){
                br = new BufferedReader(new InputStreamReader(System.in));
                out = new PrintWriter(new OutputStreamWriter(System.out));
            }
            else{
                br  = new BufferedReader(new FileReader(fileName+".in" ));
                out = new PrintWriter   (new FileWriter(fileName+".out"));
            }
        } catch (Exception e) {
            MLE();
        }
        long tb = System.currentTimeMillis();
        solve();
//        if( !oj )
//            out.println("TIME: " + (System.currentTimeMillis() - tb) / 1e3) ;
        exit(0);
    }

    String next(){
        while( st==null || !st.hasMoreElements() ){
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                return null;
            }
        }
        return st.nextToken();
    }
    String nextLine(){
        try {
            return br.readLine();
        } catch (Exception e) {
            return null;
        }
    }
    int nextInt(){ return Integer.parseInt(next()); }
    double nextDouble(){ return Double.parseDouble(next()); }
    long nextLong(){ return Long.parseLong(next()); }

    void exit( int val ){
        out.flush();
        System.exit(val);
    }

    void MLE(){
        int[][] arr = new int[1024*1024][];
        for( int i = 0; i < 1024*1024; ++i )
            arr[i] = new int[1024*1024];
    }

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









    int gcd( int a, int b ){ return b==0? a : gcd( b, a%b ); }

    void solve(){
        for( int i = 1; i<1e7; ++i )
            if( 1 < gcd(i,i+1) )
                out.println( i );


        int n, k;

        n = nextInt();
        k = nextInt();

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

        ArrayList<Integer> ans = new ArrayList<>();

        for( int i = 2; i <= n-k; ++i ) ans.add( i );
        ans.add( 1 );
        for( int i = n-k+1; i <= n  ; ++i ) ans.add( i );

        for( int v : ans )out.print( " " + v );out.println();
    }
}
