Robert is a famous engineer. One day he was given a task by his boss. The background of the task was the following:

Given a map consisting of square blocks. There were three kinds of blocks: Wall, Grass, and Empty. His boss wanted to place as many robots as possible in the map. Each robot held a laser weapon which could shoot to four directions (north, east, south, west) simultaneously. A robot had to stay at the block where it was initially placed all the time and to keep firing all the time. The laser beams certainly could pass the grid of Grass, but could not pass the grid of Wall. A robot could only be placed in an Empty block. Surely the boss would not want to see one robot hurting another. In other words, two robots must not be placed in one line (horizontally or vertically) unless there is a Wall between them.

Now that you are such a smart programmer and one of Robert‘s best friends, He is asking you to help him solving this problem. That is, given the description of a map, compute the maximum number of robots that can be placed in the map.

Input

The first line contains an integer T (<= 11) which is the number of test cases.

For each test case, the first line contains two integers m and n (1<= m, n <=50) which are the row and column sizes of the map. Then m lines follow, each contains n characters of ‘#‘, ‘*‘, or ‘o‘ which represent Wall, Grass, and Empty, respectively.

Output

For each test case, first output the case number in one line, in the format: "Case :id" where id is the test case number, counting from 1. In the second line just output the maximum number of robots that can be placed in that map.

Sample Input

2
4 4
o***
*###
oo#o
***o
4 4
#ooo
o#oo
oo#o
***#

Sample Output

Case :1
3
Case :2
5

题意：有个 n * m地图，地图由方格组成，#代表墙壁， *代表草地，o代表空地。

如今要在地图上放置机器人，每一个机器人都配备了激光枪，能够向上下左右四个方向开枪。发射的激光能够穿透草地。但不能穿透墙壁，机器人仅仅能放置在空地上。两个机器人不能放置在同一行同一列。除非他们之间有一堵墙隔开，问地图上能够放置的机器人的最大数目。

解题：这题和HDU5093基本一模一样。题中有草地，我们根本不用管，具体思路请看 ： HDU5093

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int map[51 * 50][51 * 50];
char str[51][51];
int x[51][51], x1;
int y[51][51], y1;
int used[51 * 51];
int link[51 * 51];
int n, m;

void init(){
    memset(map, 0, sizeof(map));
    memset(x, 0, sizeof(x));
    memset(y, 0, sizeof(y));
}

void creat_x(){
    x1 = 1;
    for(int i = 0; i < n; ++i){
        for(int j = 0; j < m; ++j){
            if(str[i][j] == 'o')
                x[i][j] = x1;
            if(str[i][j] == '#')
                x1++;
        }
        x1++;
    }
}

void creat_y(){
    y1 = 1;
    for(int j = 0; j < m; ++j){
        for(int i = 0; i < n; ++i){
            if(str[i][j] == 'o')
                y[i][j] = y1;
            if(str[i][j] == '#')
                y1++;
        }
        y1++;
    }
}

void getmap(){
    creat_x();
    creat_y();
    for(int i = 0; i < n; ++i){
        for(int j = 0; j < m; ++j)
            if(str[i][j] == 'o')
            map[x[i][j]][y[i][j]] = 1;
    }
}

bool dfs(int x){
    for(int i = 1; i < y1; ++i){
        if(map[x][i] && !used[i]){
            used[i] = 1;
            if(link[i] == -1 || dfs(link[i])){
                link[i] = x;
                return true;
            }
        }
    }
    return false;
}

int hungary(){
    int ans = 0;
    memset(link, -1, sizeof(link));
    for(int j = 1; j < x1; ++j){
        memset(used, 0, sizeof(used));
        if(dfs(j))
            ans++;
    }
    return ans;
}
int main (){
    int T;
    int k = 1;
    scanf("%d", &T);
    while(T--){
        init();
        scanf("%d%d", &n, &m);
        for(int i = 0; i < n; ++i)
            scanf("%s", str[i]);
        getmap();
        printf("Case :%d\n", k++);
        int sum = hungary();
        printf("%d\n", sum);
    }
    return 0;
}