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,   ...,   pn, consisting of n distinct positive integers, each is no more than n. We'll define number n as the length of permutation p1,   p2,   ...,   pn.

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, ..., a2n — the required permutation a. It is guaranteed that the solution exists. If there are multiple solutions, you can print any of them.

Sample test(s)

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.

解题说明：此题虽然是数论题，但是可以用偷懒的方法来做，首先保证前面k对数相减都为1，后面k 对数相减都为-1 。这样保证后面一项为0，前面为2k。

#include
#include
#include
using namespace std;

int main()
{
	int n, k,j;
	scanf("%d %d", &n,&k);
	for (j = 0; j<2 * k; j += 2)
	{
		printf("%d %d ", j + 2, j + 1);
	}
	for (j; j<2 * n; j += 2)
	{
		printf("%d %d ", j + 1, j + 2);
	}
	printf("\n");
	return 0;
}