Problem: Vivek has encountered a problem. He has a maze that can be represented as an n × m grid. Each of the grid cells may represent the following:

  * Empty — '.' 
  * Wall — '#' 
  * Good person — 'G' 
  * Bad person — 'B' 



The only escape from the maze is at cell (n, m).

A person can move to a cell only if it shares a side with their current cell and does not contain a wall. Vivek wants to block some of the empty cells by replacing them with walls in such a way, that all the good people are able to escape, while none of the bad people are able to. A cell that initially contains 'G' or 'B' cannot be blocked and can be travelled through.

Help him determine if there exists a way to replace some (zero or more) empty cells with walls to satisfy the above conditions.

It is guaranteed that the cell (n,m) is empty. Vivek can also block this cell.

Input

The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers n, m (1 ≤ n, m ≤ 50) — the number of rows and columns in the maze.

Each of the next n lines contain m characters. They describe the layout of the maze. If a character on a line equals '.', the corresponding cell is empty. If it equals '#', the cell has a wall. 'G' corresponds to a good person and 'B' corresponds to a bad person.

Output

For each test case, print "Yes" if there exists a way to replace some empty cells with walls to satisfy the given conditions. Otherwise print "No"

You may print every letter in any case (upper or lower).

Example

Input


6
1 1
.
1 2
G.
2 2
#B
G.
2 3
G.#
B#.
3 3
#B.
#..
GG.
2 2
#B
B.


Output


Yes
Yes
No
No
Yes
Yes

Note

For the first and second test cases, all conditions are already satisfied.

For the third test case, there is only one empty cell (2,2), and if it is replaced with a wall then the good person at (1,2) will not be able to escape.

For the fourth test case, the good person at (1,1) cannot escape.

For the fifth test case, Vivek can block the cells (2,3) and (2,2).

For the last test case, Vivek can block the destination cell (2, 2).

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import static java.lang.Math.*;
import java.util.*;
import java.io.PrintStream;
import java.io.PrintWriter;
 
public class A {
 
	static final int mod = 1000000007;
	static final int temp = 998244353;
	static final long MOD = 1000000007;
	static final long M = (long)1e9+7;
 
	
	static class Pair implements Comparable<Pair>
	{
		int first, second;
		public Pair(int aa, int bb)
		{
			first = aa; second = bb;
		}
		public int compareTo(Pair p)
		{
//			if(a == p.a) return b - p.b;
//			return a - p.a;
			if(first == p.first) return (int)(second - p.second);
			return (int)(first - p.first);
		}
	}
	
	/*
	 * IO FOR 2D GRID IN JAVA
	 * 	char[][] grid = new char[n][m];
			for(int i = 0;i<n;i++)
			{
				char[] s = sc.next().toCharArray();
				for(int j = 0;j<m;j++)
				{
					grid[i][j] = s[j];
				}
			}
	 * */
	
	static class Reader {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		int[] readArray(int n) throws IOException {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		
		long[] longReadArray(int n) throws IOException {
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
		
		
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
	}
	
 
	public static boolean isPrime(long n) {
		if(n == 1)
		{
			return false;
		}
		for(long i = 2;i*i<=n;i++)
		{
			if(n%i == 0)
			{
				return false;
			}
		}
		return true;
	}
	
	public static List<Integer> Sieve(int n)
	{
		boolean prime[] = new boolean[n+1];
		Arrays.fill(prime, true);
		
		List<Integer> l = new ArrayList<>();
		for (int p=2; p*p<=n; p++) 
		{ 
			if (prime[p] == true) 
			{ 	
			    for(int i=p*p; i<=n; i += p) 
				{
				    prime[i] = false; 
				}	
			} 
		} 
 
		for (int p=2; p<=n; p++) 
		{
		    if (prime[p] == true)
		    {
		       l.add(p); 
		    }
		}
		
		return l;
	}
	
	
	public static int gcd(int a, int b)
	{
		if(b == 0)
		 return a;
		
		else
		return gcd(b,a%b);
	} 
 
	public static int lcm(int a, int b)
    {
        return (a / gcd(a, b)) * b;
    }
	
	public static long LongGCD(long a, long b)
	{
		if(b == 0)
			 return a;
			
			else
			return LongGCD(b,a%b);
	}
	
	public static long LongLCM(long a, long b)
    {
        return (a / LongGCD(a, b)) * b;
    }
	
	public static int phi(int n)  //euler totient function
    { 
        int result = 1; 
        for (int i = 2; i < n; i++) 
            if (gcd(i, n) == 1) 
                result++; 
        return result; 
    }
	
	    
	public static int[] computePrefix(int arr[], int n)
	{
		int[] prefix = new int[n];
		prefix[0] = arr[0];
		for(int i = 1;i<n;i++)
		{
			prefix[i] = prefix[i-1]+arr[i];
		}
		
		return prefix;
	}
	
	
	public static long fastPow(long x, long n) //include mod at each step if asked and in args of fn too
	{
		if(n == 0)
			return 1;
		else if(n%2 == 0)
			return fastPow(x*x,n/2);
		else
			return x*fastPow(x*x,(n-1)/2);
	}
	
	
	   public static long modPow(long x, long y, long p)
	    {
	 
	        long res = 1;
	 
	        x = x % p;
	 
	        while (y > 0) {
	            if (y % 2 == 1)
	                res = (res * x) % p;
	 
	            y = y >> 1; 
	            x = (x * x) % p;
	        }
	 
	        return res;
	    }
	   
	   static long modInverse(long n, long p)
	    {
	        return modPow(n, p - 2, p);
	    }
	 
	    // Returns nCr % p using Fermat's little theorem.
	    
	    public static long nCr(long n, long r,long p)
	    {
	          if (n<r) 
	              return 0;
	          
	        if (r == 0)
	            return 1;
	 
	        long[] fac = new long[(int)(n) + 1];
	        fac[0] = 1;
	 
	        for (int i = 1; i <= n; i++)
	            fac[i] = fac[i - 1] * i % p;
	 
	        return (fac[(int)(n)] * modInverse(fac[(int)(r)], p)
	                % p * modInverse(fac[(int)(n - r)], p)
	                % p)
	            % p;
	    }
	    
	    public static long[] modInvArr(int n, long mod)
	    {
	    	long[] ans = new long[n+1];
			ans[1] = 1;
			for(int i = 2;i <= n;i++)
			{
				ans[i] = (mod - (mod/i) * ans[(int)(mod%i)])%mod;
			}
			return ans;
	    }
 
		public static void Sort(long[] a) {
			List<Long> l=new ArrayList<>();
			for (long i:a) l.add(i);
			Collections.sort(l);
			//Collections.reverse(l);  //Use to Sort decreasingly 
			for (int i=0; i<a.length; i++) a[i]=l.get(i);
		}
		
		public static void ssort(char[] a) {
			List<Character> l=new ArrayList<>();
			for (char ch : a) l.add(ch);
			Collections.sort(l);
			//Collections.reverse(l);  //Use to Sort decreasingly 
			for (int i=0; i<a.length; i++) a[i]=l.get(i);
		}
		
		//Modular Operations for Addition and Multiplication. 
		
		   public static long perfomMod(long x){
		        return ((x%M + M)%M);
		    }
		   public static long addMod(long a, long  b){
		        return perfomMod(perfomMod(a)+perfomMod(b));
		    }
		   public static long mulMod(long  a, long b){
		        return perfomMod(perfomMod(a)*perfomMod(b));
		   }
		
		   public static int LowerBound(int a[], int x) {
				  int l=-1,r=a.length;
				  while(l+1<r) {
				    int m=(l+r)>>>1;
				    if(a[m]>=x) r=m;
				    else l=m;
				  }
				  return r;
				}
			
			
			public static int UpperBound(int a[], int x) {
			    int l=-1;
			    int r=a.length;
			    while(l+1<r) {
			       int m=(l+r)>>>1;
			       if(a[m]<=x) l=m;
			       else r=m;
			    }
			    return l+1;
			}
			
		   
	public static void main(String[] args) throws IOException 
	{
		Reader sc=new Reader();
		PrintWriter out = new PrintWriter(System.out);
		
		int t = sc.nextInt();
	outer:	while(t-- > 0)
		{
			int n = sc.nextInt(),m = sc.nextInt();
			char[][] map = new char[n][m];
			for(int i = 0;i<n;i++)
			{
				char[] s = sc.next().toCharArray();
				for(int j = 0;j<m;j++)
				{
					map[i][j] = s[j];
				}
			}
			
			int[] dr = { 1, 0, -1, 0 };
			int[] dc = { 0, 1, 0, -1 };
			for(int i = 0;i < n;i++){
				for(int j = 0;j < m;j++){
					if(map[i][j] == 'B'){
						for(int k = 0;k < 4;k++){
							int nr = i + dr[k], nc = j + dc[k];
							if(nr >= 0 && nr < n && nc >= 0 && nc < m && map[nr][nc] == '.'){
								map[nr][nc] = '#';
							}
						}
					}
				}
			}
			
			Queue<int[]> q = new LinkedList<>();
			boolean[][] con = new boolean[n][m];
			if(map[n-1][m-1] != '#'){
				q.offer(new int[]{n-1, m-1});
				con[n-1][m-1] = true;
			}
			while(!q.isEmpty()){
				int[] cur = q.poll();
				int r = cur[0], c = cur[1];
				for(int k = 0;k < 4;k++){
					int nr = r + dr[k], nc = c + dc[k];
					if(nr >= 0 && nr < n && nc >= 0 && nc < m && map[nr][nc] != '#' && !con[nr][nc]){
						con[nr][nc] = true;
						q.offer(new int[]{nr, nc});
					}
				}
			}
			for(int i = 0;i < n;i++){
				for(int j = 0;j < m;j++){
					if(map[i][j] == 'G' && !con[i][j]){
						System.out.println("No");
						continue outer;
					}
				}
			}
			for(int i = 0;i < n;i++){
				for(int j = 0;j < m;j++){
					if(map[i][j] == 'B' && con[i][j]){
						System.out.println("No");
						continue outer;
					}
				}
			}
			System.out.println("Yes");
		}
		
		out.close();
	}
	
}