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 random
import sys

#def perm(n, k):
#result = []
data = raw_input().split()

n = int(data[0])
k = int(data[1])

if n == k:
    print -1
    #return [-1]
    sys.exit(0)
if n == 1:
    print 1
    #return [1]
    sys.exit(0)
bn = n - k - 1
swapF = (bn % 2) == 1
bn /= 2
if swapF:
    print n,
    #result.append(n)
else:
    print 1,
    #result.append(1)
complete = False
for i in xrange(2, n, 2):
    if bn == 0:
        for j in xrange(i, n):
            print j,
            #result.append(j)
        break
    if i == n - 1:
        complete = True
    print i + 1, i,
    #result.append(i + 1)
    #result.append(i)
    bn -= 1

if not complete:
    if swapF:
        print 1
        #result.append(1)
    else:
        print n
#        result.append(n)