Problem:
Rooted Trees

A graph G = (V, E) is a data structure where V is a finite set of vertices and E is a binary relation on V represented by a set of edges. Fig. 1 illustrates an example of a graph (or graphs).

Fig. 1

A free tree is a connnected, acyclic, undirected graph. A rooted tree is a free tree in which one of the vertices is distinguished from the others. A vertex of a rooted tree is called "node."

Your task is to write a program which reports the following information for each node u of a given rooted tree T:

node ID of u

parent of u

depth of u

node type (root, internal node or leaf)

a list of chidlren of u

If the last edge on the path from the root r of a tree T to a node x is (p, x), then p is the parent of x, and x is a child of p. The root is the only node in T with no parent.

A node with no children is an external node or leaf. A nonleaf node is an internal node

The number of children of a node x in a rooted tree T is called the degree of x.

The length of the path from the root r to a node x is the depth of x in T.

Here, the given tree consists of n nodes and evey node has a unique ID from 0 to n-1.

Fig. 2 shows an example of rooted trees where ID of each node is indicated by a number in a circle (node). The example corresponds to the first sample input.

Fig. 2

Input

The first line of the input includes an integer n, the number of nodes of the tree.

In the next n lines, the information of each node u is given in the following format:

id k c1 c2 ... ck

where id is the node ID of u, k is the degree of u, c1 ... ck are node IDs of 1st, ... kth child of u. If the node does not have a child, the k is 0.

Output

Print the information of each node in the following format ordered by IDs:

node id: parent = p, depth = d, type, [c1...ck]

p is ID of its parent. If the node does not have a parent, print -1.

d is depth of the node.

type is a type of nodes represented by a string (root, internal node or leaf). If the root can be considered as a leaf or an internal node, print root.

c1...ck is the list of children as a ordered tree.

Please follow the format presented in a sample output below.

Constraints

1 ≤ n ≤ 100000

Sample Input 1

13
0 3 1 4 10
1 2 2 3
2 0
3 0
4 3 5 6 7
5 0
6 0
7 2 8 9
8 0
9 0
10 2 11 12
11 0
12 0

Sample Output 1

node 0: parent = -1, depth = 0, root, [1, 4, 10]
node 1: parent = 0, depth = 1, internal node, [2, 3]
node 2: parent = 1, depth = 2, leaf, []
node 3: parent = 1, depth = 2, leaf, []
node 4: parent = 0, depth = 1, internal node, [5, 6, 7]
node 5: parent = 4, depth = 2, leaf, []
node 6: parent = 4, depth = 2, leaf, []
node 7: parent = 4, depth = 2, internal node, [8, 9]
node 8: parent = 7, depth = 3, leaf, []
node 9: parent = 7, depth = 3, leaf, []
node 10: parent = 0, depth = 1, internal node, [11, 12]
node 11: parent = 10, depth = 2, leaf, []
node 12: parent = 10, depth = 2, leaf, []

Sample Input 2

4
1 3 3 2 0
0 0
3 0
2 0

Sample Output 2

node 0: parent = 1, depth = 1, leaf, []
node 1: parent = -1, depth = 0, root, [3, 2, 0]
node 2: parent = 1, depth = 1, leaf, []
node 3: parent = 1, depth = 1, leaf, []

Note

You can use a left-child, right-sibling representation to implement a tree which has the following data:

the parent of u

the leftmost child of u

the immediate right sibling of u

Reference

Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

Solution in C (judge verdict: Time Limit Exceeded)

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define MAX 100000
/*--------構造定義---------*/
typedef struct node{
 int id;
 int depth;
 char name[20];/*root or leaf*/
 int childnum;/*子の番号を文字列で*/
 int pareid;
 struct node *point; 
}node;
/*---------------------------*/

/*--------構造定義---------*/
typedef struct CHILD{
 int childid;
 struct CHILD *point; 
}CHILD;
/*---------------------------*/

CHILD* makecell(int x){
CHILD *new=malloc(sizeof(CHILD));

new->childid=x;
new->point=NULL;

return new;
}




/*---------------depthを求める---------------*/
int depthcheck(node *p,int count,int rootid){
if(p==NULL){return count;}
else if((p->id)==rootid){count++;return count;}
else{count=1+depthcheck(p->point,count,rootid);return count;}
}
/*------------------------------------------------------------*/


int main(){
int i,j,n,d,x;
int root,count;
int depthmax=-1;
int depthmin=21;
node node[MAX];
char temp[15];
int t;
int c;
CHILD S[MAX];
CHILD **p;
CHILD *q;
CHILD *r;

scanf("%d",&n);/*node数の読み込み*/



/*-------------ノードの初期化-------------*/
for(i=0;i<n;i++){
node[i].id=i;
node[i].point=NULL;
node[i].depth=0;
node[i].childnum=0;
node[i].pareid=0;
}
/*--------------------------------------*/

/*-------------ノードの初期化-------------*/
for(i=0;i<n;i++){
 S[i].childid=-1;
 S[i].point=NULL; 
}
/*--------------------------------------*/



/*---------nodeの読み込み---------------------*/
for(i=0;i<n;i++){
scanf("%d",&t);
node[t].id=t;
scanf("%d",&node[t].childnum);

if(node[t].childnum!=0){
  for(j=0;j<node[t].childnum;j++){
   scanf("%d",&x);
   node[x].point=&node[t];
   node[x].pareid=t;
   
   if(S[t].childid==-1){
    S[t].childid=x;
   }else{
     p=&(S[t].point);
     while((*p)!=NULL){p=&((*p)->point);}
     (*p)=makecell(x);
   }

  }/*for end*/

  }
else if(node[t].childnum==0){strncpy(node[t].name,"leaf",4);node[t].name[4]='\0';}
}
/*---------------------------------------*/





/*----------root探索---------------*/
for(i=0;i<n;i++){
 if(node[i].point==NULL){root=i;node[i].pareid=-1;break;}
}
/*--------------------------------*/



/*-----------------depthの計測-----------------------*/
for(i=0;i<n;i++){
count=0;
node[i].depth=depthcheck(node[i].point,count,root);
if(node[i].depth>depthmax){depthmax=node[i].depth;}
if(node[i].depth<depthmin){depthmin=node[i].depth;}
/*--------------------------------------------------*/}




/*----------------nodeの種類の決定--------------------------*/
for(i=0;i<n;i++){
if(node[i].depth==0){strncpy(node[i].name,"root",4);node[i].name[4]='\0';}
else if(node[i].name[0]!='l'){strncpy(node[i].name,"internal node",13);node[i].name[13]='\0';}
}
/*------------------------------------------------------------*/


/*表示*/
for(i=0;i<n;i++){c=1;

printf("node %d: parent = %d, depth = %d, %s, [",i,node[i].pareid,node[i].depth,node[i].name);

if(node[i].childnum==0){printf("]\n");}/*leaf*/

else if(node[i].childnum==1){/*子が1つ*/
 printf("%d]\n",S[i].childid);
}else{/*子が2以上*/
 printf("%d, ",S[i].childid);
 q=S[i].point;
 while(q->point!=NULL){
  printf("%d, ",q->childid);
  q=q->point;
 }
  printf("%d]\n",q->childid);
 }





}



return 0;
}