Your task is to write a program which reads a rooted binary tree T and prints the following information for each node u of T:

node ID of u
parent of u
sibling of u
the number of children of u
depth of u
height of u
node type (root, internal node or leaf)
If two nodes have the same parent, they are siblings. Here, if u and v have the same parent, we say u is a sibling of v (vice versa).

The height of a node in a tree is the number of edges on the longest simple downward path from the node to a leaf.

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

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 is given in the following format:

id left right

id is the node ID, left is ID of the left child and right is ID of the right child. If the node does not have the left (right) child, the left(right) is indicated by -1.

Output
Print the information of each node in the following format:

node id: parent = p, sibling = s, degree = deg, depth = dep, height = h, type

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

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

deg, dep and h are the number of children, depth and height of the node respectively.

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.

Constraints
1 ≤ n ≤ 25
Sample Input 1
9
0 1 4
1 2 3
2 -1 -1
3 -1 -1
4 5 8
5 6 7
6 -1 -1
7 -1 -1
8 -1 -1
Sample Output 1
node 0: parent = -1, sibling = -1, degree = 2, depth = 0, height = 3, root
node 1: parent = 0, sibling = 4, degree = 2, depth = 1, height = 1, internal node
node 2: parent = 1, sibling = 3, degree = 0, depth = 2, height = 0, leaf
node 3: parent = 1, sibling = 2, degree = 0, depth = 2, height = 0, leaf
node 4: parent = 0, sibling = 1, degree = 2, depth = 1, height = 2, internal node
node 5: parent = 4, sibling = 8, degree = 2, depth = 2, height = 1, internal node
node 6: parent = 5, sibling = 7, degree = 0, depth = 3, height = 0, leaf
node 7: parent = 5, sibling = 6, degree = 0, depth = 3, height = 0, leaf
node 8: parent = 4, sibling = 5, degree = 0, depth = 2, height = 0, leaf

#include<iostream>
#include<cstdio>
using namespace std;

#define MAX 25
#define NIL -1

struct Node
{
int parent;
int left;
int right;
int h;
int depth;
int degree;
int sibling;
};

Node tree[MAX];
int root;
int n;
int max_depth=0;

int searching(int this_node, int dp)
{
tree[this_node].depth = dp;
int h1=0,h2=0, height=0;
max_depth = max(max_depth, dp);
if(tree[this_node].left!=NIL)
h1 = searching(tree[this_node].left, dp+1);
if(tree[this_node].right!=NIL)
h2 = searching(tree[this_node].right, dp+1);
if(tree[this_node].left==NIL&&tree[this_node].right==NIL)
{
tree[this_node].h = 0;
return 0;
}

height = max(h1+1, h2+1);
tree[this_node].h = height;
return height;

}

void print()
{
for(int i=0;i<n;i++)
{
printf("node %d: parent = %d, sibling = %d, degree = %d, depth = %d, height = %d, ", i, tree[i].parent, tree[i].sibling, tree[i].degree,tree[i].depth, tree[i].h);

if(tree[i].parent==NIL)
cout<<"root";
else if(tree[i].degree==0)
cout<<"leaf";
else
cout<<"internal node";
if(i!=n-1) cout<<endl;
}
}

int main()
{

for(int i=0;i<MAX;i++)
{
tree[i].parent = NIL;
tree[i].left = NIL;
tree[i].right = NIL;
tree[i].sibling = NIL;
tree[i].degree = 2;
tree[i].h = 0;
tree[i].depth = 0;
}

cin>>n;
for(int i=0;i<n;i++)
{
int current_node, left_node, right_node;
cin>>current_node>>left_node>>right_node;
tree[current_node].left = left_node;
tree[current_node].right = right_node;
tree[left_node].parent = current_node;
tree[right_node].parent = current_node;
tree[tree[current_node].left].sibling = tree[current_node].right;
tree[tree[current_node].right].sibling = tree[current_node].left;
if(left_node == NIL) tree[current_node].degree--;
if(right_node == NIL) tree[current_node].degree--;
}

for(int i=0;i<n;i++)
{
if(tree[i].parent==NIL)
{
root = i;
break;
}
}

searching(root,0);
print();

}