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
import math
def gcd(a, b):
    	c = max(a,b)
    	d = min(a,b)
    	r = c%d
    	if r==0:
    		return d
    	return gcd(d,r)
    	
def lcm(a, b):
    def gcd_naive(a, b):
    	c = max(a,b)
    	d = min(a,b)
    	r = c%d
    	if r==0:
    		return d
    	return gcd_naive(d,r)

    return int(a*b/gcd_naive(a,b))

def fn(n,k):
    l=[0]*(n)
    if n<=k:
        print(-1)
        return 0
    elif n==k+1:
        for i in range(n):
            l[i]=i+1
        print(*l)
        return 0
    else:
        for i in range(1,k+1):
            l[i]=i+1
        l[0]=n
        for i in range(n-2,k,-1):
            l[i+1]=i+1
        l[k+1]=1
        print(*l)
        
    
    
if __name__ == '__main__':    
    input = sys.stdin.read()
    data = list(map(int, input.split()))
    n = data[0]
    k = data[1]
    (fn(n,k))