Zhejiang University has 40000 students and provides 2500 courses. Now given the registered course list of each student, you are supposed to output the student name lists of all the courses.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 numbers: N (<=40000), the total number of students, and K (<=2500), the total number of courses. Then N lines follow, each contains a student's name (3 capital English letters plus a one-digit number), a positive number C (<=20) which is the number of courses that this student has registered, and then followed by C course numbers. For the sake of simplicity, the courses are numbered from 1 to K.

Output Specification:

For each test case, print the student name lists of all the courses in increasing order of the course numbers. For each course, first print in one line the course number and the number of registered students, separated by a space. Then output the students' names in alphabetical order. Each name occupies a line.

Sample Input:

10 5

ZOE1 2 4 5

ANN0 3 5 2 1

BOB5 5 3 4 2 1 5

JOE4 1 2

JAY9 4 1 2 5 4

FRA8 3 4 2 5

DON2 2 4 5

AMY7 1 5

KAT3 3 5 4 2

LOR6 4 2 4 1 5

Sample Output:

1 4

ANN0

BOB5

JAY9

LOR6

2 7

ANN0

BOB5

FRA8

JAY9

JOE4

KAT3

LOR6

3 1

BOB5

4 7

BOB5

DON2

FRA8

JAY9

KAT3

LOR6

ZOE1

5 9

AMY7

ANN0

BOB5

DON2

FRA8

JAY9

KAT3

LOR6

ZOE1

分析：

使用什么样的容器进行存储是关键。

参考：http://blog.csdn.net/sunbaigui/article/details/8656971

参考代码：

#include  
#include  
#include  
#include  
#include  
#include  
using namespace std;  
  
typedef struct Node  
{  
    char name[20];  
    bool operator < (const Node & n)const  
    {  
        return strcmp(name,n.name)<0;  
    }  
}Node;  
  
vector node;  
  
bool cmp(int t1,int  t2)  
{  
    return strcmp(node[t1].name, node[t2].name)<0;  
}  
  
  
int main()  
{  
    int N,K;  
    scanf("%d%d",&N,&K);  
    vector< vector > result(K);  
    node.resize(N); //重新分配大小，注意这一句的写法。  
    int i;  
    int j;  
    for(i=0; i