Problem: 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.

Solution (in PYTHON3):

Solution:
R = lambda :map(int,input().split())
n,k = R()
k = n-k
if k==0:
    print(-1)
    exit(0)
if n==k:
    s = list(map(str,range(1,n+1)))
    print(s[-1]+' '+' '.join(s[:-1]))
else:
    s = ''
    b1 = k//2
    b2 = k%2
    b3 = (b1 if b2==1 else b1-1)
    b = 0
    #print(b1,b2,b3)
    while b<n:
        if b==0:
            if b2==1:
                s+='1 2 '
            else:
                s+='2 1 '
            b+=2
        elif b<2*b3+2:
            s+=str(b+2)+' '+str(b+1)+' '
            b+=2
        else:
            s+=str(b+1)+' '
            b+=1
    print(s)