#include <bits/stdc++.h>
using namespace std;

const int N = 510, M = N * N * 2;
int dx[] = {0, 0, -1, 1};
int dy[] = {-1, 1, 0, 0};

// 链式前向星
int e[M], h[N], idx, w[M], ne[M];
void add(int a, int b, int c = 0) {
    e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
}

// 匈牙利算法
int st[N], match[N];
int dfs(int u) {
    for (int i = h[u]; ~i; i = ne[i]) {
        int v = e[i];
        if (st[v]) continue;
        st[v] = 1;
        if (match[v] == -1 || dfs(match[v])) {
            match[v] = u;
            return 1;
        }
    }
    return 0;
}

char g[50][50];
int main() {
#ifndef ONLINE_JUDGE
    freopen("LightOJ1152.in", "r", stdin);
#endif
    int T, n, m, cas = 0;
    scanf("%d", &T);
    while (T--) {
        // 初始化链式前向星
        memset(h, -1, sizeof h);
        idx = 0;

        scanf("%d%d", &n, &m);
        // 按char[]数组读入原始地图
        for (int i = 1; i <= n; i++) scanf(" %s", g[i] + 1);

        // 给每个金子重新编号
        int cnt = 0;
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= m; j++)
                if (g[i][j] == '*') {
                    cnt++;
                    for (int k = 0; k < 4; k++) {
                        int tx = i + dx[k], ty = j + dy[k];
                        if (g[tx][ty] != '*') continue;
                        if (tx < 1 || tx > n || ty < 1 || ty > m) continue;
                        add((i - 1) * m + j, (tx - 1) * m + ty); // 对每个金子，和它四个方向上的金子连边，没有就不连
                    }
                }
        int res = 0;
        memset(match, -1, sizeof match);
        for (int i = 1; i <= n * m; i++) {
            memset(st, 0, sizeof st);
            if (dfs(i)) res++;
        }
      
        printf("Case %d: %d\n", ++cas, cnt - res / 2);
    }
    return 0;
}