A?permutation?p?is an ordered group of numbers?p1,???p2,???...,???pn, consisting of?ndistinct 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.

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.

?

題意：找1-n的數，組成

的式子，使得最后的結果為2*k

我們可以來構造數列解決：設前2*k個數對前項比后項多1，2*k到2*n個數前項比后項小1，則式子的結果為

k+（n-k）-|k-（n-k）因為題目中說2*k<=n所以2*k符合題目要求

代碼：

#include<cstdio> #include<iostream> #include<cstring> #include<algorithm> #include<queue> #include<stack> #include<vector> #include<map> #include<set> #define MAX 300005 using namespace std;typedef long long ll; int a[MAX]; int main() {int n,k;cin>>n>>k;for(int t=1;t<=2*k;t+=2){a[t]=t;a[t+1]=t+1;}for(int t=2*k+1;t<=2*n;t+=2){a[t]=t+1;a[t+1]=t;}for(int t=1;t<=2*n;t++){cout<<a[t]<<" ";}return 0;}

?

轉載于:https://www.cnblogs.com/Staceyacm/p/10781800.html

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