#pragma GCC optimize("Ofast") 
#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define fi first
#define se second
#define sz(a) (int)(a.size())
#define all(a) a.begin(),a.end()
#define lb lower_bound
#define ub upper_bound
#define owo ios_base::sync_with_stdio(0);cin.tie(0);
#define INF (ll)(1e18)
#define debug(...) fprintf(stderr, __VA_ARGS__),fflush(stderr)
#define time__(d) for(long blockTime = 0; (blockTime == 0 ? (blockTime=clock()) != 0 : false);\
debug("%s time : %.4fs\n", d, (double)(clock() - blockTime) / CLOCKS_PER_SEC))
typedef long long int ll;
typedef long double ld;
typedef pair<ll,ll> PII;
typedef pair<int,int> pii;
typedef vector<vector<int>> vii;
typedef vector<vector<ll>> VII;
using cd = complex<ld>;
ll gcd(ll a,ll b){if(!b)return a;else return gcd(b,a%b);}
bitset<4000>a[1000];
string g[1000];
bool vis[4000];
int dx[4] = {0,0,1,-1};
int dy[4] = {1,-1,0,0};
int n,m;
void get(int i,int j){
    queue<pii>q;
    q.push({i,j});
    for(int k=0;k<4000;k++){
        a[i*m+j][k] = 0;
        vis[k] = 0;
    }
    vis[i*m+j] = 1;
    while(!q.empty()){
        int x = q.front().fi;
        int y = q.front().se;
        q.pop();
        int M = (2*m-1);
        int idx = (i-x+(n-1))*M + (j-y+(m-1));
        a[i*m+j][idx] = 1;
        for(int k=0;k<4;k++){
            int xn = x+dx[k];
            int yn = y+dy[k];
            if(xn>=0 && xn<n && yn>=0 && yn<m && g[xn][yn] == '.' && !vis[xn*m+yn]){
                q.push({xn,yn});
                vis[xn*m+yn] = 1;
            }
        } 

    }
}
int main()
{
    owo
    int t;
    cin>>t;
    while(t--){
        cin>>n>>m;
        vector<int>in(n*m,1);
        for(int i=0;i<n;i++)cin>>g[i];
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(g[i][j] == '.')get(i,j);
                else in[i*m+j] = 0;
            }
        }
        for(int i=0;i<n*m;i++){
            if(g[i/m][i%m] == 'O')continue;
            for(int j=i+1;j<n*m;j++){
                if(g[j/m][j%m] == 'O')continue;
                auto res = a[i]&a[j];
                if(res == a[i])in[i] = 0;
                if(res == a[j])in[j] = 0;
            }
        }
        int ans = 0;
        for(int i=0;i<n*m;i++)ans+=in[i];
        cout<<ans<<'\n';
    }



}