Codeforces 359B Permutation

B. Permutation

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

A permutation p is an ordered group of numbers p₁,   p₂,   ...,   p_n, consisting of n distinct positive integers, each is no more than n. We'll define number n as the length of permutation p₁,   p₂,   ...,   p_n.

Simon has a positive integer n and a non-negative integer k, such that 2k ≤ n. Help him find permutation a of length 2n, such that it meets this equation: .

Input

The first line contains two integers n and k (1 ≤ n ≤ 50000, 0 ≤ 2k ≤ n).

Output

Print 2n integers a₁, a₂, ..., a_2n — the required permutation a. It is guaranteed that the solution exists. If there are multiple solutions, you can print any of them.

Examples

Input

1 0

Output

1 2

Input

2 1

Output

3 2 1 4

Input

4 0

Output

2 7 4 6 1 3 5 8

Note

Record |x| represents the absolute value of number x.

In the first sample |1 - 2| - |1 - 2| = 0.

In the second sample |3 - 2| + |1 - 4| - |3 - 2 + 1 - 4| = 1 + 3 - 2 = 2.

In the third sample |2 - 7| + |4 - 6| + |1 - 3| + |5 - 8| - |2 - 7 + 4 - 6 + 1 - 3 + 5 - 8| = 12 - 12 = 0.

题目大意：给你一个n和一个k，要找出2*n个数满足他给的公式

首先先让所有的i满足：a[2i-1]=a[2i]+1，那么上面方程的两个求和式都应该等于N，接下来把其中某一组a[2i-1]和a[2i]交换位置，对第一个式子，很明显还是N，但是第二个减少2了，因为这个求和式某个值由1变成-1也就是说交换一组之后它们的差值是2，而方程的右边是2k，那么我们就交换k组就行了。此时第二个求和式的值是|N-2k|，由于题目限制了N>=2k，所以绝对值可以直接拿掉

#include
using namespace std;
int num[100010];
int main(){
	int n,k;
	//freopen("in.txt","r",stdin);
	while(cin>>n>>k){
		int flag=0;
		for(int i=1;i<=k;i++){
			if(!flag) printf("%d %d",i*2-1,i*2);
			else printf(" %d %d",i*2-1,i*2);
			flag=1;
		}
		for(int i=k+1;i<=n;i++){
			if(!flag) printf("%d %d",i*2,i*2-1);
			else printf(" %d %d",i*2,i*2-1);
			flag=1;
		}
		cout<