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:
//package com.example.programmingpractice;

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Scanner;

public class CodeforcesNT {

    public static void main (String[] args) throws Exception{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String[] inputs = br.readLine().trim().split(" ");
        int n = Integer.parseInt(inputs[0]);
        int k = Integer.parseInt(inputs[1]);
        int diff = n-k;
        if(diff<=0) {System.out.println("-1"); return;}
        int[] arr = new int[n+1];
        int pos = k+1;
        for(int i = 2; i<=pos; i++) arr[i] = i;
        if(diff % 2 == 1) {
            arr[1] = 1;
            pos++;
        } else {
            arr[pos+1] =  1;
            arr[1] = pos+1;
            pos+=2;
        }
        for(int i = pos; i<=n; i+=2) {
            arr[i] = i+1;
            arr[i+1] = i;
        }
        StringBuilder sb = new StringBuilder();
        for(int i = 1; i<n; i++) sb.append(arr[i]).append(' ');
        sb.append(arr[n]);
        System.out.println(sb);
    }
}