There is a company that has N employees(numbered from 1 to N),every employee in the company has a immediate boss (except for the leader of whole company).If you are the immediate boss of someone,that person is your subordinate, and all his subordinates are your subordinates as well. If you are nobody's boss, then you have no subordinates,the employee who has no immediate boss is the leader of whole company.So it means the N employees form a tree.

The company usually assigns some tasks to some employees to finish.When a task is assigned to someone,He/She will assigned it to all his/her subordinates.In other words,the person and all his/her subordinates received a task in the same time. Furthermore,whenever a employee received a task,he/she will stop the current task(if he/she has) and start the new one.

Write a program that will help in figuring out some employee’s current task after the company assign some tasks to some employee.

Input

The first line contains a single positive integer T( T <= 10 ), indicates the number of test cases.

For each test case:

The first line contains an integer N (N ≤ 50,000) , which is the number of the employees.

The following N - 1 lines each contain two integers u and v, which means the employee v is the immediate boss of employee u(1<=u,v<=N).

The next line contains an integer M (M ≤ 50,000).

The following M lines each contain a message which is either

"C x" which means an inquiry for the current task of employee x

or

"T x y"which means the company assign task y to employee x.

(1<=x<=N,0<=y<=10^9)

Output

For each test case, print the test case number (beginning with 1) in the first line and then for every inquiry, output the correspond answer per line.

Sample Input

1 
5 
4 3 
3 2 
1 3 
5 2 
5 
C 3 
T 2 1
 C 3 
T 3 2 
C 3

Sample Output

Case #1:
-1 
1 
2

解题思路：

这道题的难点就在于如何将题中的对应关系转化为线段树的模型。我们先将其给出的点的关系建立一个图（也可以称之为树，不过不是二叉树）然后从公司老板（也就是入度为0的点）出发，用dfs来确定每个点的影响范围并记录下来，在更新某个点的时候转化过来就好了，至于查询就是简单的单点查询了。

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int maxn = 50010;
int data[maxn], ss[maxn], tt[maxn], head[maxn], Tree[maxn * 4], lazy[maxn * 4];
int t, n, m, top, len;
bool vis[maxn];

 ///接下来结构体和add函数是用数组模拟邻接表，也可以换用其他的方式来存储图的信息只要能dfs遍历就行
struct Edge                          
{
	int v;
	int next;
}edge[maxn * 2];

void add(int u, int v)
{
	edge[top].v = v;
	edge[top].next = head[u];
	head[u] = top++;
}

void dfs(int x)
{
	for(int i = head[x]; i != -1; i = edge[i].next)
	{
		if(!vis[edge[i].v])
		{
			vis[edge[i].v] = true;
			ss[edge[i].v] = len + 1;///至于为什么加1，以及下边为什么++len，把样例拿出来自己试试就明白了
			dfs(edge[i].v);
			tt[edge[i].v] = ++len;
		}
	}
}

void update(int L, int R, int data, int l, int r, int rt)
{
	if(l >= L && r <= R)
	{
		Tree[rt] = data;
		lazy[rt] = data;
		return ;
	}
	if(lazy[rt] != -1)
	{
		Tree[rt << 1] = lazy[rt];
		lazy[rt << 1] = lazy[rt];
		Tree[rt << 1 | 1] = lazy[rt];
		lazy[rt << 1 | 1] = lazy[rt];
		Tree[rt] = -1;
		lazy[rt] = -1;
	}
	int m = (l + r) / 2;
	if(m >= L)
		update(L, R, data, l, m, rt << 1);
	if(m < R)
		update(L, R, data, m + 1, r, rt << 1 | 1);
}

int query(int pos, int l, int r, int rt)
{
	if(l == r && l == pos)
	{
		return Tree[rt];
	}
	if(lazy[rt] != -1)
	{
		Tree[rt << 1] = lazy[rt];
		lazy[rt << 1] = lazy[rt];
		Tree[rt << 1 | 1] = lazy[rt];
		lazy[rt << 1 | 1] = lazy[rt];
		Tree[rt] = -1;
		lazy[rt] = -1;
	}
	int m = (l + r) / 2;
	int ans = 0;
	if(m >= pos)
		ans += query(pos, l, m, rt << 1);
	else
		ans += query(pos, m + 1, r, rt << 1 | 1);
	return ans;
}

int main()
{
   // freopen("in.txt", "r", stdin);
	scanf("%d", &t);
	int cnt = 1;
	while(t--)
	{
		scanf("%d", &n);
		memset(Tree, -1, sizeof(Tree));
		memset(lazy, -1, sizeof(lazy));
		memset(vis, false, sizeof(vis));
		memset(head, -1, sizeof(head));
		top = 0;
		len = 0;
		int a, b;
		for(int i = 1; i <= n - 1; ++ i)
		{
			scanf("%d%d", &a, &b);
			add(b, a);
			vis[a] = true;
		}
		int k;
		for(int i = 1; i <= n; ++ i) ///找到最终boss
		{
			if(!vis[i])
			{
				k = i;
				break;
			}
		}
		memset(vis, false, sizeof(vis)); ///重新初始化用来标记是否走过
		vis[k] = true;
		ss[k] = 1;
		dfs(k);
		tt[k] = ++len;
		scanf("%d", &m);
		char st[4];
		int x, y;
		printf("Case #%d:\n", cnt++);
		for(int i = 1; i <= m; ++ i)
		{
			scanf("%s", st);
			if(st[0] == 'C')
			{
				scanf("%d", &x);
				//cout << "query " << tt[x] << endl;
				printf("%d\n", query(tt[x], 1, n, 1));
			}
			else
			{
				scanf("%d%d",  &x, &y);
				//cout << "update " << ss[x] << ' ' <<tt[x] << endl;
				update(ss[x], tt[x], y, 1, n, 1);
			}
		}
	}
	return 0;
}