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.
# coding: utf-8

"""
Questão 361B: Levko and Permutation.
Leonardo de C. Cordeiro - UFCG
lcxcordeiro@gmail.com
"""

import fractions

n, k = map(int, raw_input().split())

if k >= n:
    print "-1"
else:

    used = [0] * n
    permutation = [0] * n

    for i in xrange(k):
        permutation[n - 1 - i] = (n - i)
        used[n - 1 - i] = 1

    len = n - k

    if len % 2 == 0:
        for i in xrange(len-1, 0, -2):
            permutation[i - 1] = i + 1
            permutation[i] = i
    else:
        permutation[0] = 1
        for i in xrange(len-1, 1, -2):
            permutation[i - 1] = i + 1
            permutation[i] = i

    result = ""
    for x in xrange(n):
        result += "%s " % permutation[x]

    print result[:-1]
