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 sys
from collections import defaultdict, Counter
from itertools import permutations, combinations
from math import sin, cos, asin, acos, tan, atan, pi

sys.setrecursionlimit(10 ** 6)

def pyes_no(condition, yes = "YES", no = "NO", none = "-1") :
  if condition == None:
    print (none)
  elif condition :
    print (yes)
  else :
    print (no)

def plist(a, s = ' ') :
  print (s.join(map(str, a)))

def rint() :
  return int(sys.stdin.readline())

def rstr() :
  return sys.stdin.readline().strip()


def rints() :
  return map(int, sys.stdin.readline().split())

def rfield(n, m = None) :
  if m == None :
    m = n
  
  field = []
  for i in xrange(n) :
    chars = sys.stdin.readline().strip()
    assert(len(chars) == m)
    field.append(chars)
  return field

def pfield(field, separator = '') :
  print ('\n'.join(map(lambda x: separator.join(x), field)))

def check_field_equal(field, i, j, value) :
  if i >= 0 and i < len(field) and j >= 0 and j < len(field[i]) :
    return value == field[i][j]
  return None 

def digits(x, p) :
  digits = []
  while x > 0 :
    digits.append(x % p)
    x //= p
  return digits[::-1]

def undigits(x, p) :
  value = 0
  for d in x :
    value *= p
    value += d
  return value

def modpower(a, n, mod) :
  r = a ** (n % 2)
  if n > 1 :
    r *= modpower(a, n // 2, mod) ** 2
  return r % mod

def gcd(a, b) :
  if a > b :
    a, b = b, a
  
  while a > 0 :
    a, b = b % a, a

  return b

def vector_distance(a, b) :
  diff = vector_diff(a, b)
  
  return scalar_product(diff, diff) ** 0.5

def vector_inverse(v) :
  r = [-x for x in v]

  return tuple(r)

def vector_diff(a, b) :
  return vector_sum(a, vector_inverse(b))

def vector_sum(a, b) :
  r = [c1 + c2 for c1, c2 in zip(a, b)]
    
  return tuple(r)

def scalar_product(a, b) :
  r = 0
  for c1, c2 in zip(a, b) :
    r += c1 * c2

  return r

def check_rectangle(points) :
  assert(len(points) == 4)

  A, B, C, D = points

  for A1, A2, A3, A4 in [
    (A, B, C, D),
    (A, C, B, D),
    (A, B, D, C),
    (A, C, D, B),
    (A, D, B, C),
    (A, D, C, B),
  ] :
    sides = (
      vector_diff(A1, A2),
      vector_diff(A2, A3),
      vector_diff(A3, A4),
      vector_diff(A4, A1),
    )
    if all(scalar_product(s1, s2) == 0 for s1, s2 in zip(sides, sides[1:])) :
       return True
  return False

def check_square(points) :
  if not check_rectangle(points) :
    return False
  A, B, C, D = points

  for A1, A2, A3, A4 in [
    (A, B, C, D),
    (A, C, B, D),
    (A, B, D, C),
    (A, C, D, B),
    (A, D, B, C),
    (A, D, C, B),
  ] :
    side_lengths = [
      (first[0] - next[0]) ** 2 + (first[1] - next[1]) ** 2 for first, next in zip([A1, A2, A3, A4], [A2, A3, A4, A1])
    ]
    if len(set(side_lengths)) == 1 :
      return True
    
  return False

def check_right(p) :
  # Check if there are same points
  for a, b in [
    (p[0], p[1]),
    (p[0], p[2]),
    (p[1], p[2]),
  ] :
    if a[0] == b[0] and a[1] == b[1] :
      return False

  a, b, c = p
  a, b, c = vector_diff(a, b), vector_diff(b, c), vector_diff(c, a)   

  return scalar_product(a, b) * scalar_product(a, c) * scalar_product(b, c) == 0

def modmatrixproduct(a, b, mod) :
  n, m1 = len(a), len(a[0])
  m2, k = len(b), len(b[0])

  assert(m1 == m2)
  m = m1

  r = [[0] * k for i in range(n)]
  for i in range(n) :
    for j in range(k) :
      for l in range(m) :
        r[i][j] += a[i][l] * b[l][j]
      r[i][j] %= mod
  return r

def modmatrixpower(a, n, mod) :
  magic = 2
  for m in [2, 3, 5, 7] :
    if n % m == 0 :
      magic = m
      break

  r = None
  if n < magic : 
    r = a
    n -= 1
  else :
    s = modmatrixpower(a, n // magic, mod)
    r = s
    for i in range(magic - 1) :
      r = modmatrixproduct(r, s, mod)

  for i in range(n % magic) : 
    r = modmatrixproduct(r, a, mod)
  
  return r

n, k = rints()
if k == n :
  print -1
else : 
  values = [n - k] + range(1, n - k) + range(n - k + 1, n + 1)

  plist(values)
