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 PYTHON3):

Solution:
import math
#import math 
#------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
 
BUFSIZE = 8192
 
 
class FastIO(IOBase):
    newlines = 0
 
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
 
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
 
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)
 
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
 
#-------------------game starts now----------------------------------------------------import math
class Graph: 
    def __init__(self): 
        self.graph = defaultdict(list) 
      
    # add edge to graph 
    def addEdge(self, u, v): 
        self.graph[u].append(v) 
  
    # BFS function to find path from source to sink      
    def BFS(self,a): 
          
        # Base case 
        # Mark all the vertices as not visited  
        visited = [False]*(n*m + 1) 
        queue = [] 
        queue.append(n*m) 
  
        # Mark the current node as visited and  
        # enqueue it  
        visited[n*m] = True
        while(queue): 
  
            # Dequeue a vertex from queue 
            s = queue.pop(0) 
  
            # Get all adjacent vertices of the  
            # dequeued vertex s. If a adjacent has  
            # not been visited, then mark it visited  
            # and enqueue it  
            for i in self.graph[s]: 
                  
                # If this adjacent node is the destination  
                # node, then return true  
  
                # Else, continue to do BFS  
                if visited[i] == False: 
                    queue.append(i) 
                    visited[i] = True
  
        for i in range(len(a)):
            if visited[a[i]]==False:
                return False
        return True
  
def isSafe(i, j, matrix): 
    if i >= 0 and i < len(matrix) and j >= 0 and j < len(matrix[0]) and matrix[i][j]!='#': 
        return True
    else: 
        return False
  
# Returns true if there is a path from a source (a  
# cell with value 1) to a destination (a cell with  
# value 2) 
def findPath(M): 
    #s, d = None, None # source and destination  
    N = len(M)
    m=len(M[0])
    g = Graph() 
  
    # create graph with n * n node  
    # each cell consider as node  
    k = 1
    a=[]# Number of current vertex 
    for i in range(N): 
        for j in range(m): 
            if (M[i][j] != '#'): 
  
                # connect all 4 adjacent cell to  
                # current cell  
                if (isSafe(i, j + 1, M)): 
                    g.addEdge(k, k + 1) 
                if (isSafe(i, j - 1, M)): 
                    g.addEdge(k, k - 1) 
                if (isSafe(i + 1, j, M)): 
                    g.addEdge(k, k + m) 
                if (isSafe(i - 1, j, M)): 
                    g.addEdge(k, k - m) 
                if (M[i][j]=='G'):
                    a.append(k)
            
            k += 1
    
    return g.BFS(a)
for i in range(int(input())):
 n,m=map(int,input().split())
 matrix=[[0 for i in range(m)]for j in range(n)]
 from collections import defaultdict    
 s=[]
 c=0
 f=0
 c1=0
 w=[]
 for j in range(n):
     s.append(list(input()))
     c+=s[-1].count('G')
 if c==0:
     print("Yes")
     continue
 for i in range(n):
     for j in range(m):
         if s[i][j]=='B':
             w.append((i,j))
 for y in w:
             i=y[0]
             j=y[1]
             if i>0:
                 if s[i-1][j]=='G':
                     f=1
                     break
                 s[i-1][j]='#'
             if j>0:
                 if s[i][j-1]=='G':
                     f=1
                     break
                 s[i][j-1]='#'
             if i<n-1:
                 if s[i+1][j]=='G':
                     f=1
                     break
                 s[i+1][j]='#'
             if j<m-1:
                 if s[i][j+1]=='G':
                     f=1
                     break
                 s[i][j+1]='#'
     
 e=[] 
 if f==1:
     print("No")
     continue
 else:
     if findPath(s)==True:
         print("Yes")
     else:
            print("No")