#include <bits/stdc++.h>

using namespace std;

struct Node
{
    int ans[4];
};

void solve()
{
    int n, m;
    cin >> n >> m;
    vector<string> g(n);

    for(int i = 0;i < n; i ++)      cin >> g[i];

    int dx[] = {0, 0, 1, -1}, dy[] = {1, -1, 0, 0};

    vector<vector<int>> block(n, vector<int> (m));//每个格子属于哪个连通块
    int cnt = 0;//连通块的数量
    
    vector<int> idx(1010);
    auto bfs = [&] (int x, int y, int id)
    {
        queue<pair<int, int>> q;
        q.push({x, y});
        block[x][y] = id;

        while(q.size())
        {
            auto [ix, iy] = q.front();
            q.pop();
            idx[id] ++;

            for(int i = 0;i < 4; i ++)
            {
                int a = dx[i] + ix, b = dy[i] + iy;
                if(a < 0 || a >= n || b < 0 || b >= m)        
                    continue;
                
                if(block[a][b] || g[a][b] == 'O')     continue;

                q.push({a, b});
                block[a][b] = id;
            }
        }
    };
    
    for(int i = 0;i < n; i ++)
        for(int j = 0;j < m; j ++)
            if(!block[i][j] && g[i][j] != 'O')
                bfs(i, j, ++ cnt);

    set<pair<int, int>> S[cnt + 1];

    for(int i = 0;i < n; i ++)
        for(int j = 0;j < m; j ++)
            S[block[i][j]].insert({i, j});
        
    for(int i = 1;i <= cnt; i ++)
    {
        int sx = -1, sy = -1;
        set<pair<int, int>> v;
        for(auto &[x, y] : S[i])
        {
            if(sx == -1 && sy == -1)
                sx = x, sy = y, v.insert({0, 0});
            else    
                v.insert({x - sx, y - sy});
        }
        S[i] = v;
    }

    int ans = 0;

    auto check = [&] (set<pair<int, int>> A, set<pair<int, int>> B)
    {
        //A能被B包含
        for(auto [x, y] : A)
            if(!B.count({x, y}))
                return false;
        
        return B.size() >= A.size();
    };

    // for(int i = 1;i <= cnt; i ++)
    //     for(auto [x, y] : S[cnt])
    //         clog << x << ' ' << y << ' ' << i << '\n';

    for(int i = 1;i <= cnt; i ++)
    {
        bool ok = true;
        for(int j = 1;j <= cnt; j ++)
        {
            if(i != j)
            {
                if(check(S[i], S[j]))
                {
                    ok = false;
                    break;
                }
            }
        }
        if(ok)  
        {
            ans += idx[i];
            //clog << i << '\n';
        }
    }

    cout << ans << '\n';

}           

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int tt;
    cin >> tt;
    while(tt --)
    {
        solve();
    }
    return 0;
}