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.
tamanho, n_bonitos = map(int, input().split(" "))
n_permuta = tamanho - n_bonitos
minha_permuta = []

ponteiro_de_troca = 1
if n_bonitos < tamanho:
    for i in range(tamanho):
        minha_permuta.append(i+1)

    if n_permuta % 2 != 0:
        ponteiro_de_troca = 2
        n_permuta -= 1

    while n_permuta > 0:
        minha_permuta[ponteiro_de_troca],minha_permuta[ponteiro_de_troca -1] = minha_permuta[ponteiro_de_troca -1],minha_permuta[ponteiro_de_troca]
        ponteiro_de_troca += 2
        n_permuta -= 2

    print(" ".join(map(str,minha_permuta)))
else:
    print(-1)
 			 	   		 		 	 	      		 			