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:
######################################################
############Created by Devesh Kumar###################
#############devesh1102@gmail.com####################
##########For CodeForces(Devesh1102)#################
#####################2020#############################
######################################################
import sys
input = sys.stdin.readline
 #qn given an arr of size n for every  i position find a min j such that j>i and arr[j]>arr[i]
#it prints the position of the nearest .
# import sys
import heapq 
import copy
import math
#heapq.heapify(li) 
#
#heapq.heappush(li,4) 
#
#heapq.heappop(li)
#
# &	Bitwise AND Operator	10 & 7 = 2
# |	Bitwise OR Operator	10 | 7 = 15
# ^	Bitwise XOR Operator	10 ^ 7 = 13

# <<	Bitwise Left Shift operator	10<<2 = 40
# >>	Bitwise Right Shift Operator
'''############ ---- Input Functions ---- #######Start#####'''
def inp():
	return(int(input()))
def inlt(): 
	return(list(map(int,input().split())))
def insr():
	s = input()
	return(list(s[:len(s) - 1]))
def invr():
	return(map(int,input().split()))
 ############ ---- Input Functions ---- #######End
 # #####   

class Node:
    def _init_(self,val):
        self.data = val
        self.left = None
        self.right = None
##to initialize wire : object_name= node(val)##to create a new node
## can also be used to create linked list


def pre_sum(arr):
	#"""returns the prefix sum inclusive ie ith position in ans  represent sum from 0 to ith position"""
	p = [0]
	for i in arr:
		p.append(p[-1] + i)
	p.pop(0)
	return p

def pre_back(arr):
	#"""returns the prefix sum inclusive ie ith position in ans  represent sum from 0 to ith position"""
	p = [0]
	for i in arr:
		p.append(p[-1] + i)
	p.pop(0)
	return p
def bin_search(arr,l,r,val):#strickly greater
	if arr[r] <= val:
		return r+1	
	if r-l < 2:
		if arr[l]>val:
			return l
		else:
			return r
	mid = int((l+r)/2)
	if arr[mid] <= val:
		return bin_search(arr,mid,r,val)
	else:
		return bin_search(arr,l,mid,val)
def search_leftmost(arr,val):
	def helper(arr,l,r,val):
		# print(arr)
		print(l,r)
		if arr[l] == val:
			return l
		if r -l <=1:
			if arr[r] == val:
				return r
			else:
				print("not found")
				return
		mid = int((r+l)/2)
		if arr[mid] >= val:
			return helper(arr,l,mid,val)
		else:
			return helper(arr,mid,r,val)
	return helper(arr,0,len(arr)-1,val)
def search_rightmost(arr,val):
	def helper(arr,l,r,val):
		# print(arr)
		print(l,r)
		if arr[r] == val:
			return r
		if r -l <=1:
			if arr[l] == val:
				return r
			else:
				print("not found")
				return
		mid = int((r+l)/2)
		if arr[mid] > val:
			return helper(arr,l,mid,val)
		else:
			return helper(arr,mid,r,val)
	return helper(arr,0,len(arr)-1,val)


def pr_list(a):
	print(*a, sep=" ")
def main():
	tests =  inp()
	# tests = 1
	mod = 1000000007

	limit = 10**18
	for test in range(tests):
		[n,m] = inlt()
		maze = []
		for i in range(n):
			maze.append(insr())
		# 	maze[-1] = ['#'] + maze[-1] + ['#'] 
		# maze = []
		good = {}
		flag = 0
		for i in range(n):
			for j in range(m):
				if maze[i][j] == "G":
					good[(i,j)] = 1
				elif maze[i][j] == "B":
					if i + 1 < n :
						if maze[i+1] [j]== "G":
							flag = 1
						elif maze[i+1][j] != "B":
							maze[i+1] [j] = "#"

					if j + 1 < m :
						if maze[i] [j+1]== "G":
							flag = 1
						elif maze[i][j+1] != "B":
							maze[i] [j+1] = "#"

					if i - 1 >= 0 :
						if maze[i-1] [j]== "G":
							flag = 1
						elif maze[i-1][j] != "B":
							maze[i-1] [j] = "#"

					if j - 1 >= 0 :
						if maze[i] [j-1]== "G":
							flag = 1
						elif maze[i][j-1] != "B":
							maze[i] [j-1] = "#"	
		if good =={}:
			print("Yes")
			continue
		if flag == 1 or maze[n-1][m-1] == "#" or maze[n-1][m-1] == "B" :
			print("No")
			continue
		flag = 0
		queue = [(n-1,m-1)]
		done = [[0 for i in range(m)] for j in range(n)]
		done [-1][-1] =1
		if (n-1,m-1) in good:
			del good[(n-1,m-1)]
		while(queue!= []):
			# print(queue)
			curr = queue.pop(0)
			i = curr[0]
			j = curr[1]
			if (i,j) in good:
				del good[(i,j)]
			if i-1 >= 0 and  maze[i-1][j] != "#" and done[i-1][j] ==0:
				queue.append((i-1,j))
				done[i-1][j] = 1
			if j-1 >= 0 and  maze[i][j-1] != "#" and done[i][j-1] ==0 :
				queue.append((i,j-1))
				done[i][j-1] = 1
			if i+1 < n and  maze[i+1][j] != "#" and done[i+1][j] ==0:
				queue.append((i+1,j))
				done[i+1][j] =1
			if j+1 < m and maze[i][j+1] != "#" and done[i][j+1] ==0:
				queue.append((i,j+1))
				done[i][j+1] = 1
		if good != {}:
			print("No")
		else:
			print("Yes")
			

			

					

		

		# print(x_values)
		
		# for i in range(x):
		# 	if int(x_values[i])!=1:
		# 		n_new = i+1 + (n-i-1)*int(x_values[i])
		# 		n = n_new%mod
		# print(n)
				
			
		




if __name__== "__main__":
  main()