There is an apple tree outside of kaka's house. Every autumn, a lot of apples will grow in the tree. Kaka likes apple very much, so he has been carefully nurturing the big apple tree.

The tree has N forks which are connected by branches. Kaka numbers the forks by 1 to N and the root is always numbered by 1. Apples will grow on the forks and two apple won't grow on the same fork. kaka wants to know how many apples are there in a sub-tree, for his study of the produce ability of the apple tree.

The trouble is that a new apple may grow on an empty fork some time and kaka may pick an apple from the tree for his dessert. Can you help kaka?

The first line contains an integer N (N ≤ 100,000) , which is the number of the forks in the tree.
The following N - 1 lines each contain two integers u and v, which means fork u and fork v are connected by a branch.
The next line contains an integer M (M ≤ 100,000).
The following M lines each contain a message which is either
"C x" which means the existence of the apple on fork x has been changed. i.e. if there is an apple on the fork, then Kaka pick it; otherwise a new apple has grown on the empty fork.
or
"Q x" which means an inquiry for the number of apples in the sub-tree above the fork x, including the apple (if exists) on the fork x
Note the tree is full of apples at the beginning

  1 ///建树+树状数组（stl）8080k 1047ms
  2 /*
  3 #include <cstdio>
  4 #include <iostream>
  5 #include <cstring>
  6 #include <vector>
  7 #include <algorithm>
  8 #define INF 0x3f3f3f3f
  9 using namespace std;
 10 typedef vector<int>ve;
 11
 12 const int MAXN  = 1e5+10;
 13 int t[MAXN], l[MAXN], r[MAXN];
 14 bool vis[MAXN];
 15 int N, M, cnt;
 16
 17 vector<ve>node(MAXN);
 18
 19 int lowbit(int x)
 20 {
 21     return x&(-x);
 22 }
 23 void add(int x, int value)
 24 {
 25     for(int i = x; i <= N; i+=lowbit(i))
 26         t[i]+=value;
 27 }
 28 int sum(int x)
 29 {
 30     int res = 0;
 31     for(int i = x; i > 0; i-=lowbit(i))
 32         res+=t[i];
 33     return res;
 34 }
 35 void dfs(int k)
 36 {
 37     l[k] = cnt;
 38     for(int i = 0; i < node[k].size(); i++)
 39     {
 40         cnt++;
 41         dfs(node[k][i]);
 42     }
 43     r[k] = cnt;
 44     return;
 45 }
 46 int main()
 47 {
 48     scanf("%d", &N);
 49     int a, b;
 50     for(int i = 1; i <= N; i++)
 51     {
 52         vis[i] = 1;
 53         add(i, 1);
 54     }
 55     for(int i = 1; i < N; i++)
 56     {
 57         scanf("%d%d", &a, &b);
 58         node[a].push_back(b);
 59     }
 60     cnt = 1;
 61     dfs(1);
 62     scanf("%d", &M);
 63     while(M--)
 64     {
 65         char com[3];
 66         int px;
 67         scanf("%s", &com);
 68         if(com[0] == 'C')
 69         {
 70             scanf("%d", &px);
 71             if(vis[px]) add(l[px], -1);
 72             else add(l[px], 1);
 73             vis[px] = !vis[px];
 74         }
 75         else
 76         {
 77             scanf("%d", &px);
 78             int ans = sum(r[px]) - sum(l[px]-1);
 79             printf("%d\n", ans);
 80         }
 81     }
 82     return 0;
 83 }
 84 */
 85
 86 ///建树+树状数组（静态连接表） 4940k 454ms
 87 #include <cstdio>
 88 #include <iostream>
 89 #include <cstring>
 90 #include <vector>
 91 #include <algorithm>
 92 #define INF 0x3f3f3f3f
 93 using namespace std;
 94
 95 const int MAXN = 1e5+10;
 96 const int MAXM = 1e5+10;
 97
 98 struct node
 99 {
100     int to, next;
101 }edge[MAXM<<1];
102
103 int head[MAXN];
104 bool vis[MAXN];
105 int t[MAXN], l[MAXN], r[MAXN];
106 int N, M, cnt, key;
107
108 int lowbit(int x)
109 {
110     return x&(-x);
111 }
112 void add(int x, int value)
113 {
114     for(int i = x; i <= N; i+=lowbit(i))
115         t[i]+=value;
116 }
117 int sum(int x)
118 {
119     int res = 0;
120     for(int i = x; i > 0; i-=lowbit(i))
121         res+=t[i];
122     return res;
123 }
124
125 void dfs(int k)
126 {
127     l[k] = key;
128     for(int i = head[k]; i != -1; i = edge[i].next)
129     {
130         if(l[edge[i].to]) continue;
131         key++;
132         dfs(edge[i].to);
133     }
134     r[k] = key;
135 }
136
137 void init()
138 {
139     memset(head, -1, sizeof(head));
140     cnt = 1;
141 }
142
143 void add_e(int u, int v)
144 {
145     edge[cnt].to = v;
146     edge[cnt].next = head[u];
147     head[u] = cnt++;
148 }
149
150 int main()
151 {
152     scanf("%d", &N);
153     init();
154     int a, b;
155     for(int i = 1; i <= N; i++)
156     {
157         vis[i] = 1;             //每个节点一开始都有苹果
158         add(i, 1);              //初始化树状数组
159     }
160     for(int i = 1; i < N; i++)
161     {
162         scanf("%d%d", &a, &b); ///建无向图
163         add_e(a, b);
164         add_e(b, a);
165     }
166     key = 1; dfs(1);           //建树
167     scanf("%d", &M);
168     int px;
169     char com[3];
170     while(M--)
171     {
172         scanf("%s", &com);
173         if(com[0] == 'C')
174         {
175             scanf("%d", &px);
176             if(vis[px]) add(l[px], -1);
177             else add(l[px], 1);
178             vis[px] = !vis[px];
179         }
180         else
181         {
182             scanf("%d", &px);
183             printf("%d\n", sum(r[px])-sum(l[px]-1));
184         }
185     }
186     return 0;
187 }