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 division, print_function


def main():
    # Template 1.0
    import sys, re, math
    from collections import deque, defaultdict, Counter, OrderedDict
    from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians, floor
    from heapq import heappush, heappop, heapify, nlargest, nsmallest
    def STR():
        return list(input())

    def INT():
        return int(input())

    def MAP():
        return map(int, input().split())

    def LIST():
        return list(map(int, input().split()))

    def list2d(a, b, c):
        return [[c] * b for i in range(a)]

    def sortListWithIndex(listOfTuples, idx):
        return (sorted(listOfTuples, key=lambda x: x[idx]))

    def sortDictWithVal(passedDic):
        temp = sorted(passedDic.items(), key=lambda kv: (kv[1], kv[0]))
        toret = {}
        for tup in temp:
            toret[tup[0]] = tup[1]
        return toret

    def sortDictWithKey(passedDic):
        return dict(OrderedDict(sorted(passedDic.items())))

    INF = float('inf')
    mod = 10 ** 9 + 7

    def apply(c):
        toCheck = []
        x, y = c[0], c[1]
        X = [-1, 1, 0, 0]
        Y = [0, 0, 1, -1]

        for i in range(4):
            if(x+X[i]>=0 and x+X[i]<n and y+Y[i]>=0 and y+Y[i]<m):
                if(grid[x + X[i]][y + Y[i]]=='G'):
                    return False
                if(grid[x + X[i]][y + Y[i]]=='.'):
                    grid[x + X[i]][y + Y[i]] = '#'
        return True

    def dfs(g):
        x, y = g[0], g[1]

        visited = list2d(n, m, False)
        X = [-1, 1, 0, 0]
        Y = [0, 0, 1, -1]

        q = deque([(x, y)])

        while(q):
            popped = q.popleft()
            i,j = popped[0], popped[1]

            if(i==n-1 and j==m-1):
                return True


            if(grid[i][j]=='G'):
                visGoods.add((i, j))

            for di in range(4):
                if(i+X[di]>=0 and i+X[di]<n and j+Y[di]>=0 and j+Y[di]<m and grid[i+X[di]][j+Y[di]]!='#' and visited[i+X[di]][j+Y[di]]==False):
                    q.append((i+X[di], j+Y[di]))
                    visited[i + X[di]][j + Y[di]] = True

        return False

    def dfs1(g):
        x, y = g[0], g[1]

        visited = list2d(n, m, False)
        X = [-1, 1, 0, 0]
        Y = [0, 0, 1, -1]

        q = deque([(x, y)])

        while (q):
            popped = q.popleft()
            i, j = popped[0], popped[1]

            if (i == n - 1 and j == m - 1):
                return True



            if (grid[i][j] == 'B'):
                visBads.add((i, j))

            for di in range(4):
                if (i + X[di] >= 0 and i + X[di] < n and j + Y[di] >= 0 and j + Y[di] < m and grid[i + X[di]][j + Y[di]] != '#' and visited[i + X[di]][j + Y[di]] == False):
                    q.append((i + X[di], j + Y[di]))
                    visited[i + X[di]][j + Y[di]] = True
        return False

    t = INT()

    while (t != 0):
        n,m = MAP()

        grid = []

        for _ in range(n):
            grid.append(list(input()))

        bads = []
        goods = []
        for i in range(n):
            for j in range(m):
                if(grid[i][j]=='B'):
                    bads.append((i, j))
                if(grid[i][j]=='G'):
                    goods.append((i, j))

        #build walls
        test = True
        for coord in bads:
            test = apply(coord)

        if(test):
            #check good persons
            ans = True
            visGoods = set()
            for g in goods:
                if(g not in visGoods):
                    ans&=dfs(g)
                    if(ans==False):
                        break

            if(ans==True):
                #check bad persons
                visBads = set()
                for b in bads:
                    if(b not in visBads):
                        if(dfs1(b)==True):
                            ans = False
                            break

            # print(visBads, visGoods)
            if(ans==False):
                print("No")
            else:
                print("Yes")
        else:
            print("No")

        t -= 1


######## Python 2 and 3 footer by Pajenegod and c1729

# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.

# So on cf, use PyPy2 for best string performance.

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')

# Cout implemented in Python
import sys


class ostream:
    def __lshift__(self, a):
        sys.stdout.write(str(a))
        return self


cout = ostream()
endl = '\n'


# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero=0):
    conv = ord if py2 else lambda x: x
    A = [];
    numb = zero;
    sign = 1;
    i = 0;
    s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'0'[0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-'[0]:
                sign = -1
            elif s[i] != b'\r'[0]:
                A.append(sign * numb)
                numb = zero;
                sign = 1
            i += 1
    except:
        pass
    if s and s[-1] >= b'0'[0]:
        A.append(sign * numb)
    return A


if __name__ == "__main__":
    main()