Problem Description

"The Werewolves" is a popular card game among young people.In the basic game, there are 2 different groups: the werewolves and the villagers.

Each player will debate a player they think is a werewolf or not.

Their words are like "Player x is a werewolf." or "Player x is a villager.".

What we know is :

1. Villager won't lie.

2. Werewolf may lie.

Of cause we only consider those situations which obey the two rules above.

It is guaranteed that input data exist at least one situation which obey the two rules above.

Now we can judge every player into 3 types :

1. A player which can only be villager among all situations,

2. A player which can only be werewolf among all situations.

3. A player which can be villager among some situations, while can be werewolf in others situations.

You just need to print out the number of type-1 players and the number of type-2 players.

No player will talk about himself.

 

Input

The first line of the input gives the number of test cases T.Then T test cases follow.

The first line of each test case contains an integer N,indicating the number of players.

Then follows N lines,i-th line contains an integer x and a string S,indicating the i-th players tell you,"Player x is a S."

limits:

1≤T≤10

1≤N≤100,000

1≤x≤N

S∈ {"villager"."werewolf"}

 

Output

For each test case,print the number of type-1 players and the number of type-2 players in one line, separated by white space.

 

Sample Input

 

Sample Output

 

Source

 #include <bits/stdc++.h>

 #define INF 0x3f3f3f3f

 #define LL long long

 using namespace std;

 const int MAXN = 1e5+;

 int N, M;

 int w[MAXN];

 vector<int>fa[MAXN], son[MAXN];

 bool vis[MAXN];

 int ans[MAXN], cnt;

 void add1(int a, int b)

 {

     fa[a].push_back(b);

 }

 bool dfs(int a, int b)

 {

     bool flag = true;

     for(int i = ; i < fa[a].size(); i++){

         if(fa[a][i] == b) return false;

         flag = dfs(fa[a][i], b);

         if(!flag) return false;

     }

     return flag;

 }

 void dfs2(int x)

 {

     if(!vis[x]) return;

     vis[x] = false;

     for(int i = ; i < fa[x].size(); i++){

         if(vis[fa[x][i]]){

             cnt++;

             dfs2(fa[x][i]);

         }

     }

 }

 int main()

 {

     //freopen("6370in.txt", "r", stdin);

     int T_case, id, tot = ;

     char ty[];

     scanf("%d", &T_case);

     while(T_case--){

         cnt = ;

         tot = ;

         memset(vis, , sizeof(vis));

         scanf("%d", &N);

         for(int i = ; i <= N; i++){ fa[i].clear();son[i].clear();}

         for(int i = ; i <= N; i++){

             scanf("%d %s", &id, &ty);

             if(ty[] == 'w'){

                 w[++tot] = i;

                 son[i].push_back(id);

             }

             else{

                 add1(id, i);

             }

         }

         for(int i = ; i <= tot; i++){

             if(!vis[w[i]]) continue;

             for(int k = ; k < son[w[i]].size();k++){

                 if(!vis[son[w[i]][k]]) continue;

                 if(!dfs(w[i], son[w[i]][k])){

                     //puts("zjy");

                     cnt++;

                     dfs2(son[w[i]][k]);

                 }

             }

         }

         printf("0 %d\n", cnt);

     }

 }