Levko and Permutation CodeForces - 361B(思维)

 

Levko and Permutation

CodeForces - 361B

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 p_i of permutation p₁, p₂, ... , p_n is good if gcd(i, p_i) > 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.

Example

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.

直接暴力的话肯定会超时，思维题，如果n和k是相等的一定是不可能的，因为如果排列是1 2 3 4 ……n最多是n-1个，1和1 最大公因数为1不行，所以如果题目要求k个，那就从最后取k个，让他的下标对应相等就行了也就是n-k+1~n 一共k个，然后把n-k这个数提到第一个，这样1~n-k-1这几个数依次后移一位，那么公因数一定为1.

code：

#include <iostream>
#include <cstdio>
using namespace std;
int main(){
	int n,k;
	cin >> n >> k;
	if(n == k)
	 cout << "-1" << endl;
	else{
		cout << n-k;
		int i;
		for(i = 1; i < n-k; i++)
		   cout << " " << i;
		for(i = n-k+1; i <= n; i++)
		   cout << " " << i;
		cout << endl;
	}
	return 0;
}