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: Accepted)

#include <stdio.h>
#include <stdlib.h>
#define ll long long
#define rep(i,l,r)for(ll i=(l);i<(r);i++)

//辺の情報を個別に持つタイプ
typedef struct edge{ll s,g;}E;
typedef struct graph{
	int vcnt,ecnt;
	E  e[200010];//適宜変える
	int id[100010];//適宜変える
}G;

int esort(const void*a,const void*b){
	E*p=(E*)a,*q=(E*)b;
	if((*p).s<(*q).s)return -1;
	if((*p).s>(*q).s)return  1;
	return 0;
}

int cnt[100010];
G g;
void readgraph(){
	//適宜変える
	ll n;
	scanf("%lld",&n);
	int m=0;
	rep(i,0,n){
		ll x,y,k;
		scanf("%lld%lld",&x,&k);
		while(k--){
			scanf("%lld",&y);
			cnt[y]++;
			g.e[m].s=x;
			g.e[m].g=y;
			m++;
			g.e[m].s=y;
			g.e[m].g=x;
			m++;
		}
	}
	g.vcnt=n;
	g.ecnt=m;
	qsort(g.e,g.ecnt,sizeof(E),esort);

	int p=0;
	rep(i,0,g.vcnt){
		while(p<g.ecnt&&g.e[p].s<i)p++;
		g.id[i]=p;
	}
	g.id[g.vcnt]=g.ecnt;//番兵
}



//dfs
int a[100010];
ll dfs(int v,int pre){
	rep(i,g.id[v],g.id[v+1]){
		if(g.e[i].g!=pre){
			//hoge
			a[g.e[i].g]=a[v]+1;
			dfs(g.e[i].g,v);
		}
	}
}

int main(){
	readgraph();
	int r;
	rep(i,0,g.vcnt)if(cnt[i]==0)r=i;
	dfs(r,-1);
	rep(i,0,g.vcnt){
		int t=-1;
		rep(j,g.id[i],g.id[i+1])if(a[g.e[j].g]<a[i]){
			t=g.e[j].g;
			break;
		}
		
		printf("node %d: parent = %d, depth = %d, ",i,t,a[i]);
		
		if(i==r)printf("root, ");
		else if(g.id[i]+1==g.id[i+1])printf("leaf, ");
		else printf("internal node, ");
		
		printf("[");
		int flag=0;
		rep(j,g.id[i],g.id[i+1])if(a[g.e[j].g]>a[i])printf(flag++?", %d":"%d",g.e[j].g);
		puts("]");
	}
}