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:
#from __future__ import print_function
mod=int(1e9+7)
#import resource
#resource.setrlimit(resource.RLIMIT_STACK, [0x100000000, resource.RLIM_INFINITY])
#import threading
#threading.stack_size(2**25)
#import sys
#sys.setrecursionlimit(10**5)
#fact=[1]
#for i in range(1,100001):
#    fact.append((fact[-1]*i)%mod)
#ifact=[0]*100001
#ifact[100000]=pow(fact[100000],mod-2,mod)
#for i in range(100000,0,-1):
#    ifact[i-1]=(i*ifact[i])%mod
#from random import randint as rn
import bisect
from bisect import bisect_left as bl              #c++ lowerbound bl(array,element)
from bisect import bisect_right as br             #c++ upperbound
import itertools
import collections
import math
import heapq
from types import GeneratorType
def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to
    return wrappedfunc
def modinv(n,p):
    return pow(n,p-2,p)
def ncr(n,r,p):                        #for using this uncomment the lines calculating fact and ifact
    t=((fact[n])*((ifact[r]*ifact[n-r])%p))%p
    return t
def GCD(x,y):
    while(y):
        x, y = y, x % y
    return x
def ain():                           #takes array as input
    return list(map(int,sin().split()))
def sin():
    return input()
def val(x):
    if(x<=0):
        return 0
    return (x*(x+1))/2
range=xrange
"""*******************************************************"""
def main():
    @bootstrap
    def dfs(x,y):
        vis[x][y]=1
        if(a[x][y]=="G"):
            s[0]+=1
        if(a[x+1][y]!="#" and vis[x+1][y]==0):
            yield(dfs(x+1,y))
        if(a[x-1][y]!="#" and vis[x-1][y]==0):
            yield(dfs(x-1,y))
        if(a[x][y+1]!="#" and vis[x][y+1]==0):
            yield(dfs(x,y+1))
        if(a[x][y-1]!="#" and vis[x][y-1]==0):
            yield(dfs(x,y-1))
        yield 0
    for _ in range(int(sin())):
        n,m=ain()
        a=[["#"]*(m+2)]
        for i in range(n):
            x=list(sin())
            x.insert(0,"#")
            x.append("#")
            a.append(x)
        a.append(["#"]*(m+2))
        f=0
        g=0
        for i in range(1,n+1):
            for j in range(1,m+1):
                if(a[i][j]=="G"):
                    g+=1
                if(a[i][j]=="B"):
                    if(a[i][j+1]=="G"):
                        f=1
                        break
                    if(a[i][j+1]!="B"):
                        a[i][j+1]="#"
                    if(a[i][j-1]=="G"):
                        f=1
                        break
                    if(a[i][j-1]!="B"):
                        a[i][j-1]="#"
                    if(a[i+1][j]=="G"):
                        f=1
                        break
                    if(a[i+1][j]!="B"):
                        a[i+1][j]="#"
                    if(a[i-1][j]=="G"):
                        f=1
                        break
                    if(a[i-1][j]!="B"):
                        a[i-1][j]="#"
            if(f==1):
                break
        if(f==1):
            print"No"
            continue
        s=[0]
        vis=[[0 for i in range(m+2)] for i in range(n+2)]
        if(a[n][m]!="#"):
            dfs(n,m)
        if(s[0]==g):
            print"Yes"
        else:
            print"No"
######## Python 2 and 3 footer by Pajenegod and c1729
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange

import os, sys
from io import IOBase, BytesIO

BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s
    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()

    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            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')
if __name__ == '__main__':
   main()
#threading.Thread(target=main).start()