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:
n,k=map(int,input().split())
if n<=k:
    print(-1)
elif k==0 and n==1:
    print(1)
elif n==1:
    print(-1)
else:
    a=[]
    l=0
    p=n-k 
    if p%2==0:
        for i in range(n):
            if i<=n-1-k:
                if i==0:
                    a.append(n-1-k+1)
                elif i==n-1-k:
                    a.append(1)
                elif i%2==0:
                    a.append(i)
                elif i%2!=0:
                    a.append(i+2)
            else:
                a.append(i+1)
    else:
        for i in range(n):
            if i<=n-1-k:
                if i==0:
                    a.append(1)
                
                elif i%2==0:
                    a.append(i)
                elif i%2!=0:
                    a.append(i+2)
            else:
                a.append(i+1)
    print(*a)