Description：

Solution：

Code：

#include<algorithm>

#include<iostream>

#include<cstdlib>

#include<cstdio>

#include<cmath>

#include<queue>

#include<cctype>

#include<cstring>

using namespace std;

#define MAXN 32

char c[32][32];

char getc()

{

	register char c = getchar();

	while(c != '.' && c != '#')c = getchar();

	return c;

}

int n,m;

int to(int i,int j){return (i - 1) * m + j;}

int s,t;

int mx[4] = {0,0,1,-1};

int my[4] = {1,-1,0,0};

struct edge

{

	int to,nxt,f;

}e[MAXN * MAXN * 4 * 2];

int edgenum = 0;

int lin[MAXN * MAXN];

void add(int a,int b,int f)

{

	e[edgenum].to = b;e[edgenum].f = f;e[edgenum].nxt = lin[a];lin[a] = edgenum;++edgenum;

	e[edgenum].to = a;e[edgenum].f = 0;e[edgenum].nxt = lin[b];lin[b] = edgenum;++edgenum;

	return;

}

int ch[MAXN * MAXN];

bool BFS()

{

	memset(ch,-1,sizeof(ch));ch[s] = 0;

	queue<int> q;q.push(s);

	while(!q.empty())

	{

		int k = q.front();q.pop();

		for(int i = lin[k];i != -1;i = e[i].nxt)

		{

			if(ch[e[i].to] == -1 && e[i].f)

			{

				ch[e[i].to] = ch[k] + 1;

				q.push(e[i].to);

			}

		}

	}

	return (ch[t] != -1);

}

int flow(int k,int f)

{

	if(k == t)return f;

	int r = 0;

	for(int i = lin[k];i != -1 && f > r;i = e[i].nxt)

	{

		if(ch[e[i].to] == ch[k] + 1 && e[i].f)

		{

			int l = flow(e[i].to,min(e[i].f,f - r));

			e[i].f -= l;r += l;e[i ^ 1].f += l;

		}

	}

	if(r == 0)ch[k] = -1;

	return r;

}

#define INF 0x3f3f3f3f

int dinic()

{

	int ans = 0,r;

	while(BFS())while(r = flow(s,INF))ans += r;

	return ans;

}

void work()

{

	scanf("%d%d",&n,&m);

	for(int i = 1;i <= n;++i)

		for(int j = 1;j <= m;++j)

			c[i][j] = getc();

	int cnt = 0,sum = 0;

	memset(lin,-1,sizeof(lin));

	edgenum = 0;

	s = n * m + 1;t = s + 1;

	for(int i = 1;i <= n;++i)

	{

		for(int j = 1;j <= m;++j)

		{

			if(c[i][j] == '#')continue;

			if((i + j) % 2 == 0)

			{

				++cnt;

				add(s,to(i,j),2);

				for(int k = 0;k < 4;++k)

				{

					if(c[i + mx[k]][j + my[k]] != 0 && c[i + mx[k]][j + my[k]] != '#')

					{

						add(to(i,j),to(i + mx[k],j + my[k]),1);

					}

				}

			}

			else

			{

				++sum;

				add(to(i,j),t,2);

			}

		}

	}

	if(cnt != sum)

	{

		puts("NO");

		return;

	}

	int res = dinic();

	if(res == cnt * 2)puts("YES");

	else puts("NO");

	return;

}

int main()

{

	int testcases;

	scanf("%d",&testcases);

	while(testcases--)work();

	return 0;

}