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.util.StringTokenizer;


public class LevkoandPermutation {
public static void main(String[] args) throws IOException {
    BufferedReader bf=  new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st = new StringTokenizer(bf.readLine());
    int n = Integer.parseInt(st.nextToken());
    int k = Integer.parseInt(st.nextToken());
    if(n==k)
        System.out.println(-1);
    else{
        int [] a = new int[n+1];
        int x;
        if(n%2==0){
        x = 2;
        for (int i = 1; i < a.length;i++) {
            a[i]=x;
            if(i%2!=0)
                x--;
            else
                x+=3;
        }
        
        
        if(k%2!=0){
            int tmp = a[1];
            a[1] = a[2];
            a[2] = tmp;
            k--;
        }
        int index = 3;
        while(k>0){
            int tmp = a[index];
            a[index] = a[index+1];
            a[index+1] = tmp;
            index+=2;
            k-=2;
        }
        for (int i = 1; i < a.length; i++) {
            System.out.print(a[i]+" ");
        }
        }
        else{
            x=3;
            a[1]=1;
            for (int i = 2; i < a.length; i++) {
                a[i] = x;
                if(i%2==0)
                    x--;
                else
                    x+=3;
            }
            if(k%2!=0){
                a[1] = n;
                a[n] = 1;
                a[n-1] = n-1;
                k--;
            }
            int index = 2;
            while(k>0){
                int tmp = a[index];
                a[index] = a[index+1];
                a[index+1] = tmp;
                index+=2;
                k-=2;
            }
            for (int i = 1; i < a.length; i++) {
                System.out.print(a[i]+" ");
            }
        }
    }
}
}
