Permutation(思维题)

Description

A permutationp 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.

Sample Input

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

Sample Output

Hint

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.

题解：因为两者差为2k，就要想办法满足这个条件，假设前2k项满足前后两项和为1，则另2k+1到2n项前后两项和为-1；

那么公式一的和为k+k,公式二为k-k，和就为2k。

代码如下：

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
int main()
{
    int n,k;
    cin>>n>>k;
    for(int i=1;i<=k;i++)
	cout<<2*i<<" "<<2*i-1<<" ";
    for(int i=k+1;i<=n;i++)
	cout<<2*i-1<<" "<<2*i<<" ";
    return 0;
}