#include<bits/stdc++.h>
using namespace std;
//#define map unordered_map
#define re register
#define ll long long
#define cll const ll
#define forl(i,a,b) for(re ll (i)=(a);i<=(b);(i)++)
#define forr(i,a,b) for(re ll (i)=(a);i>=(b);(i)--)
#define forll(i,a,b,c) for(re ll (i)=(a);i<=(b);(i)+=(c))
#define forrr(i,a,b,c) for(re ll (i)=(a);i>=(b);(i)-=(c))
#define forL(i,a,b,c) for(re ll (i)=(a);((i)<=(b)) && (c);(i)++)
#define forR(i,a,b,c) for(re ll (i)=(a);((i)>=(b)) && (c);(i)--)
#define forLL(i,a,b,c,d) for(re ll (i)=(a);((i)<=(b)) && (d);(i)+=(c))
#define forRR(i,a,b,c,d) for(re ll (i)=(a);((i)>=(b)) && (d);(i)-=(c))
#define pii pair<ll,ll>
#define mid ((l+r)>>1)
#define lowbit(x) (x&-x)
#define pb push_back
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
#define endl '\n'
#define QwQ return 0;
#define db long double
#define ull unsigned long long
#define lcm(x,y) (1ll*(x)/__gcd(x,y)*(y))
#define Sum(x,y) (1ll*((x)+(y))*((y)-(x)+1)/2)
#define x first
#define y second
#define aty cout<<"Yes\n";
#define atn cout<<"No\n";
#define cfy cout<<"YES\n";
#define cfn cout<<"NO\n";
#define xxy cout<<"yes\n";
#define xxn cout<<"no\n";
#define printcf(x) x?cout<<"YES\n":cout<<"NO\n";
#define printat(x) x?cout<<"Yes\n":cout<<"No\n";
#define printxx(x) x?cout<<"yes\n":cout<<"no\n";
#define maxqueue priority_queue<ll>
#define minqueue priority_queue<ll,vector<ll>,greater<ll>>
#define bug cout<<"---------------------------------------\n";
ll pw(ll x,ll y,ll mod){if(y==0)return 1;if(x==0)return 0;ll an=1,tmp=x;while(y){if(y&1)an=(an*tmp)%mod;tmp=(tmp*tmp)%mod;y>>=1;}return an;}
template<typename T1,typename T2>bool Max(T1&x,T2 y){if(y>x)return x=y,1;return 0;}template<typename T1,typename T2>bool Min(T1&x,T2 y){if(y<x)return x=y,1;return 0;}
cll mod=1e9+7,N=300010;ll fac[N],inv[N];void init(){fac[0]=1;forl(i,1,N-5)fac[i]=i*fac[i-1],fac[i]%=mod;inv[N-5]=pw(fac[N-5],mod-2,mod);forr(i,N-5,1)inv[i-1]=inv[i]*i%mod;}ll C(ll n,ll m){if(n<m || m<0)return 0;return (fac[n]%mod)*(inv[n-m]*inv[m]%mod)%mod;}
ll _t_;
/*
12
2 5
.OO..
O..O.
1 3
O.O
1 3
.O.
2 3
OOO
OOO
3 3
O..
...
...
2 3
.O.
.O.
2 3
.O.
...
3 6
OOO..O
O..O.O
OOOOOO
2 2
.O
O.
2 3
..O
OO.
3 3
O.O
.O.
O..
3 3
O.O
.O.
O.O

3
1
0
0
8
0
5
3
0
2
3
0

对于每个袋鼠，找它的连通块 -> O(nm)。

对于每个袋鼠，找他的连通块所有的能走的点 -> O(nm)  

这样是 O(n^3m^3) 的，考虑优化。

发现每个袋鼠能干掉自己连通块的所有袋鼠，没了？

~~似乎是 O(n^2m^2) 的。 ~~

貌似极限数据 2s，怎么回事呢。 
我擦？有个结论是不是一个连通块中一个袋鼠能赢，所有袋鼠能赢啊？ 
*/
void _clear(){}
ll n,m;
ll k;
char a[1010][1010];
vector<pii>G[1010];
ll vis[1010][1010];
ll dx[]={0,0,1,-1},dy[]={1,-1,0,0};
queue<pii>q;
bool check(ll x,ll y){
	return x>=1 && x<=n && y>=1 && y<=m && a[x][y]=='.';
}
ll VIS[1010][1010];
vector<pii>now,rel;
ll ans;
void upd(ll addx,ll addy)
{
	vector<pii>nw;
	forl(i,0,(ll)rel.size()-1)
		if(check(rel[i].x+addx,rel[i].y+addy))
			nw.pb(rel[i]);
	rel=nw;
}
void solve()
{
	ans=0;
	while(!q.empty())
		q.pop();
	k=0;
    _clear();
	cin>>n>>m;
	forl(i,1,n)
		forl(j,1,m)
			cin>>a[i][j];
	forl(i,0,n+2)
		forl(j,0,m+2)
			vis[i][j]=0;
	forl(i,0,n*m+5)
		G[i].clear();
	forl(i,1,n)
		forl(j,1,m)
			if(a[i][j]=='.' && !vis[i][j])
			{
				vis[i][j]=++k;
				G[k].pb({i,j});
				q.push({i,j});
				while(!q.empty())
				{
					pii now=q.front();
					q.pop();
					forl(l,0,3)
					{
						ll fx=dx[l]+now.x,fy=dy[l]+now.y;
						if(!vis[fx][fy] && check(fx,fy))
							vis[fx][fy]=k,
							G[k].pb({fx,fy}),
							q.push({fx,fy});
					}
				}
			}
/*	forl(i,1,n)
	{
		forl(j,1,m)
			cout<<vis[i][j]<<' ';
		cout<<endl;
	}*/
	forl(i,1,k)
	{
		now.clear();
		forl(j,1,k)
			if(i!=j)
				forl(l,0,(ll)G[j].size()-1)
					now.pb(G[j][l]);
		forl(J,0,(ll)G[i].size()-1)
		{
			rel=now;
			pii j=G[i][J];
			forl(K,0,(ll)G[i].size()-1)
			{
				pii l=G[i][K];
				ll addx=l.x-j.x,addy=l.y-j.y;
				upd(addx,addy);
			}
	//		cout<<j.x<<' '<<j.y<<' '<<rel.size()<<endl;
			if(rel.size()==0)
				ans+=(ll)G[i].size();
			break;
		}
	}
	cout<<ans<<endl;
}
int main()
{
    IOS;
    _t_=1;
    cin>>_t_;
    while(_t_--)
        solve();
    QwQ;
}