Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K = 3, then you must output 3→2→1→6→5→4; if K = 4, you must output 4→3→2→1→5→6.

Input Specification:

Each input file contains one test case. For each case, the first line contains the address of the first node, a positive N (<= 10⁵) which is the total number of nodes, and a positive K (<=N) which is the length of the sublist to be reversed. The address of a node is a 5-digit nonnegative integer, and NULL is represented by -1.

Then N lines follow, each describes a node in the format:

Address Data Next

where Address is the position of the node, Data is an integer, and Next is the position of the next node.

Output Specification:

For each case, output the resulting ordered linked list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input:

00100 6 4

00000 4 99999

00100 1 12309

68237 6 -1

33218 3 00000

99999 5 68237

12309 2 33218

Sample Output:

00000 4 33218

33218 3 12309

12309 2 00100

00100 1 99999

99999 5 68237

68237 6 -1

 #include <cstdio>

 #include <cstring>

 #include <string>

 #include <queue>

 #include <stack>

 #include <cmath>

 #include <iostream>

 #include <algorithm>

 using namespace std;

 struct node{

     int now,next,data;

 };

 node mem[];

 vector<node> v;

 int main(){

     //freopen("D:\\INPUT.txt","r",stdin);

     int h,now,data,next,num,k,temp;

     scanf("%d %d %d",&h,&num,&k);

     int i;

     for(i=;i<num;i++){

         scanf("%d %d %d",&now,&data,&next);

         mem[now].now=now;

         mem[now].data=data;

         mem[now].next=next;

     }

     now=h;

     while(now!=-){

         v.push_back(mem[now]);

         now=mem[now].next;

     }

     int length=v.size();

     int round=length/k;

     for(i=;i<round;i++){

         reverse(v.begin()+i*k,v.begin()+i*k+k);

     }

     for(i=;i<length-;i++){

         printf("%05d %d %05d\n",v[i].now,v[i].data,v[i+].now);

     }

     printf("%05d %d %d\n",v[i].now,v[i].data,-);//注意全反的情况

     return ;

 }