CodeForces

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.

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.

题意：给定n和k，找到一个长度为n的序列，里面的数是1~n且洽有k个数和1的gcd大于1，其余n-k个数和1的gcd均等于1。

思路：1和任何数的gcd都等于1，除1外任何数和它自己的gcd都大于1，和它下一个数的gcd都等于1。

不可能所有的数的gcd都大于1（因为有1）。

下边 2~k+1 为数本身（gcd大于1）

k+2~n 为下标加一，即k+3~n+1%n=1（gcd等于1）

1 为k+2

#include <iostream>#include <cstdio>#include <algorithm>#define max_ 3010using namespace std;int n,k;int main(){cin>>n>>k;if(n==k){printf("-1\n");return 0;}for(int i=1;i<=n;i++){if(i==1)printf("%d",(k+2-1)%n+1);else if(i-1<=k)printf(" %d",i);elseprintf(" %d",(i+1-1)%n+1);}printf("\n");}