B. Levko and Permutation

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Sample test(s)

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.

解题说明：此题的意思是给定一个n，得到一个从1到n的数列。现在随机排练该数列，求出数列中的每个数字和其下标位置的最大公约数，如果这个公约数大于1就记为幸运数，现在给定一个k，确保幸运数的个数正好为k。这里可以用构造法，为了得到k个幸运数，我们可以让后k个数字正好与其位置上面的数字相同，然后让前n-k个数按相反的顺序排列，这样就能保证只有k个幸运数。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include <cstring>
using namespace std;

int main()
{
	int n, k, i;
	scanf("%d%d", &n, &k);
	if (n == k)
	{
		printf("-1\n");
	}
	else
	{
		printf("%d ", n - k);
		for (i = 1; i < n - k; i++)
		{
			printf("%d ", i);
		}
		for (i = n - k + 1; i <=n; i++)
		{
			printf("%d ", i);
		}
		printf("\n");
	}
	return 0;
}