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.
def checkKey(dict, key): 
	if key in dict:
		return True
	return False
# def helper(s):
# 	l=len(s)
# 	if (l==1):
# 		l=[]
# 		l.append(s)
# 		return l
# 	ch=s[0]
# 	recresult=helper(s[1:])
# 	myresult=[]
# 	myresult.append(ch)
# 	for st in recresult:
# 		myresult.append(st)
# 		ts=ch+st
# 		myresult.append(ts)

# 	return myresult


# mod=1000000000+7
# def helper(s,n,open,close,i):
# 	if(i==2*n):
# 		for i in s:
# 			print(i,end='')
# 		print()
# 		return
# 	if(open<n):
# 		s[i]='('
# 		helper(s,n,open+1,close,i+1)
# 	if(close<open):
# 		s[i]=')'
# 		helper(s,n,open,close+1,i+1)

# def helper(arr,i,n):
# 	if(i==n-1):
# 		recresult=[arr[i]]
# 		return recresult
# 	digit=arr[i]
# 	recresult=helper(arr,i+1,n)
# 	myresult=[]
# 	for i in recresult:
# 		myresult.append(i)
# 		myresult.append(i+digit);
# 	myresult.append(digit)
# 	return myresult
# import copy
# n=int(input())
# arr=list(map(int,input().split()))
# ans=[]
# def helper(arr,i,n):
# 	if(i==n-1):
# 		# for a in arr:
# 		# 	print(a,end=" ")
# 		# print()
# 		l=copy.deepcopy(arr)
# 		ans.append(l)
# 		return
# 	for j in range(i,n):
# 		if(i!=j):
# 			if(arr[i]==arr[j]):
# 				continue
# 			else:
# 				arr[i],arr[j]=arr[j],arr[i]
# 				helper(arr,i+1,n)
# 				arr[j],arr[i]=arr[i],arr[j]
# 		else:
# 			arr[i],arr[j]=arr[j],arr[i]
# 			helper(arr,i+1,n)
# def helper(sol,n,m):
# 	for i in range(n+1):
# 		for j in range(m+1):
# 			print(sol[i][j],end=" ")
# 		print()
# def rat_in_a_maze(maze,sol,i,j,n,m):
# 	if(i==n and j==m):
# 		sol[i][j]=1
# 		helper(sol,n,m)
# 		exit()
# 	if(i>n or j>m):
# 		return False
# 	if(maze[i][j]=='X'):
# 		return False
# 	maze[i][j]='X'
# 	sol[i][j]=1
# 	if(rat_in_a_maze(maze,sol,i,j+1,n,m)):
# 		return True	
# 	elif(rat_in_a_maze(maze,sol,i+1,j,n,m)):
# 		return True
# 	sol[i][j]=0
# 	return False
n,k=map(int,input().split())
if(k<=n-1):
	l=[1]
	for i in range(1,n+1):
		l.append(i)
	index=n-k
	while(index-1>=1):
		l[index],l[index-1]=l[index-1],l[index]
		index=index-1
	for i in range(1,n+1):
		print(l[i],end=" ")
	print()



else:
	print(-1)





