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 PYTHON):

Solution:
"""
    Satwik_Tiwari ;) .
    28 june , 2020  - Sunday
"""

#===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function

from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase


import bisect
from heapq import *
from math import *
from collections import deque
from collections import Counter as counter  # Counter(list)  return a dict with {key: count}
from itertools import combinations as comb # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
#If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
#If the element is already present in the list,
# the right most position where element has to be inserted is returned

#==============================================================================================

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")


def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()


if sys.version_info[0] < 3:
    sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

# inp = lambda: sys.stdin.readline().rstrip("\r\n")

#===============================================================================================
#some shortcuts

mod = 1000000007
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var))  #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
# def graph(vertex): return [[] for i in range(0,vertex+1)]
def zerolist(n): return [0]*n
def nextline(): out("\n")  #as stdout.write always print sring.
def testcase(t):
    for p in range(t):
        solve()
def printlist(a) :
    for p in range(0,len(a)):
        out(str(a[p]) + ' ')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(a,b):
    ans = 1
    while(b>0):
        if(b%2==1):
            ans*=a
        a*=a
        b//=2
    return ans
def ncr(n,r): return factorial(n)//(factorial(r)*factorial(max(n-r,1)))
def isPrime(n) : # Check Prime Number or not
    if (n <= 1) : return False
    if (n <= 3) : return True
    if (n % 2 == 0 or n % 3 == 0) : return False
    i = 5
    while(i * i <= n) :
        if (n % i == 0 or n % (i + 2) == 0) :
            return False
        i = i + 6
    return True

#===============================================================================================
# code here ;))
def bs(a,l,h,x):
    while(l<h):
        # print(l,h)
        mid = (l+h)//2
        if(a[mid] == x):
            return mid
        if(a[mid] < x):
            l = mid+1
        else:
            h = mid
    return l

def sieve(a): #O(n loglogn) nearly linear
    #all odd mark 1
    for i in range(3,((10**6)+1),2):
        a[i] = 1
    #marking multiples of i form i*i 0. they are nt prime
    for i in range(3,((10**6)+1),2):
        for j in range(i*i,((10**6)+1),i):
            a[j] = 0
    a[2] = 1 #special left case
    return (a)


def bfs(g,st):
    visited = [-1]*(len(g))
    visited[st] = 0
    queue = []
    queue.append(st)
    new = []
    while(len(queue) != 0):
        s = queue.pop()
        new.append(s)
        for i in g[s]:
            if(visited[i] == -1):
                visited[i] = visited[s]+1
                queue.append(i)
    return visited


def dfsusingstack(v,st):
    d = deque([])
    visited = [0]*(len(v))
    d.append(st)
    new = []
    visited[st] = 1
    while(len(d) != 0):
        curr = d.pop()
        new.append(curr)
        for i in v[curr]:
            if(visited[i] == 0):
                visited[i] = 1
                d.append(i)
    return new


def solve():
    n,m = sep()
    grid = []
    for i in range(n):
        temp = list(inp())
        grid.append(temp)

    cnt  = 0
    for i in range(n):
        for j in range(m):
            if(grid[i][j] == 'B'):
                if(i!=0):
                    if(grid[i-1][j] == '.'):
                        grid[i-1][j] = '#'
                    elif(grid[i-1][j] == 'G'):
                        print('No')
                        return
                if(i!=n-1):
                    if(grid[i+1][j] == '.'):
                        grid[i+1][j] = '#'
                    elif(grid[i+1][j] == 'G'):
                        print('No')
                        return
                if(j!= 0):
                    if(grid[i][j-1] == '.'):
                        grid[i][j-1] = '#'
                    elif(grid[i][j-1] == 'G'):
                        print('No')
                        return
                if(j!=m-1):
                    if(grid[i][j+1] == '.'):
                        grid[i][j+1] = '#'
                    elif(grid[i][j+1] == 'G'):
                        print('No')
                        return

    # print(grid)
    good = [False]*(n*m)
    cnt = 0
    for i in range(0,n):
        for j in range(0,m):
            if(grid[i][j] == '.'):
                grid[i][j] = cnt
                cnt+=1
                continue
            if(grid[i][j] == 'G'):
                grid[i][j] = cnt
                good[cnt] = True
                cnt+=1
                continue

    if(grid[n-1][m-1] == '#'):
        if(good.count(True) == 0):
            print('Yes')
        else:
            print('No')
        return

    graph = [[] for i in range(cnt)]
    for i in range(n):
        for j in range(m):
            if(grid[i][j] != "#" and grid[i][j] != 'B'):
                if(i!=0 and grid[i-1][j] != "#" and grid[i-1][j] != 'B'):
                    graph[grid[i][j]].append(grid[i-1][j])
                    # graph[grid[i-1][j]].append(grid[i][j])
                if(i!=n-1 and grid[i+1][j] != "#" and grid[i+1][j] != 'B'):
                    graph[grid[i][j]].append(grid[i+1][j])
                    # graph[grid[i+1][j]].append(grid[i][j])
                if(j!=0 and grid[i][j-1] != "#" and grid[i][j-1] != 'B'):
                    graph[grid[i][j]].append(grid[i][j-1])
                    # graph[grid[i][j-1]].append(grid[i][j])
                if(j!= m-1 and grid[i][j+1] != "#" and grid[i][j+1] != 'B'):
                    graph[grid[i][j]].append(grid[i][j+1])
                    # graph[grid[i][j+1]].append(grid[i][j])
    # print(graph)

    # print(good)
    traversal = dfsusingstack(graph,cnt-1)
    # print(traversal)
    cnt = 0
    for i in range(len(traversal)):
        if(good[traversal[i]]):
            cnt+=1
    if(cnt == good.count(True)):
        print('Yes')
    else:
        print('No')



# testcase(1)
testcase(int(inp()))