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 bisect
import heapq
import math
import collections
import sys
import copy
from functools import reduce
import decimal
from io import BytesIO, IOBase
import os

sys.setrecursionlimit(10 ** 9)
decimal.getcontext().rounding = decimal.ROUND_HALF_UP

graphDict = collections.defaultdict
queue = collections.deque


################## Graphs ###################
class Graphs:
    def __init__(self):
        self.graph = graphDict(list)

    def add_edge(self, u, v):
        self.graph[u].append(v)
        self.graph[v].append(u)

    def dfs_utility(self, nodes, visited_nodes):
        visited_nodes.add(nodes)
        print(nodes, end="")
        for neighbour in self.graph[nodes]:
            if neighbour not in visited_nodes:
                self.dfs_utility(neighbour, visited_nodes)

    def dfs(self, node):
        Visited = set()
        self.dfs_utility(node, Visited)

    def bfs(self, node):
        visited = set()
        if node not in visited:
            queue.append(node)
            visited.add(node)
        while queue:
            parent = queue.popleft()
            print(parent)
            for item in self.graph[parent]:
                if item not in visited:
                    queue.append(item)
                    visited.add(item)


#################### Tree Implementaion ##############
class Tree:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None


def inorder(node):
    if node:
        inorder(node.left)
        print(node.data)
        inorder(node.right)


def leaf_node_sum(root):
    if root is None:
        return 0
    if root.left is None and root.right is None:
        return root.data
    return leaf_node_sum(root.left) + leaf_node_sum(root.right)


def hight(root):
    if root is None:
        return -1
    if root.left is None and root.right is None:
        return 0
    return max(hight(root.left), hight(root.right)) + 1


#################################################

def rounding(n):
    return int(decimal.Decimal(f'{n}').to_integral_value())


def factors(n):
    return set(reduce(list.__add__,
                      ([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0)))


################################ <fast I/O> ###########################################
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)


#############################################<I/O Region >##############################################


def inp():
    return sys.stdin.readline().strip()


def map_inp(v_type):
    return map(v_type, inp().split())


def list_inp(v_type):
    return list(map_inp(v_type))


######################################## Solution ####################################
def flood_fill(r, c, color):
    visit = [(r, c)]
    while visit:
        rr, cc = visit.pop()
        if rr < 0 or rr > n - 1 or cc < 0 or cc > m - 1 or arr[rr][cc] == "#" or arr[rr][cc] == "G" or visited[rr][cc]:
            continue
        if arr[rr][cc] == ".":
            arr[rr][cc] = "#"
            continue
        else:
            visited[rr][cc] = True
            visit.append((rr + 1, cc))
            visit.append((rr - 1, cc))
            visit.append((rr, cc + 1))
            visit.append((rr, cc - 1))


def bfs(r, c):
    global count
    visit = queue([(r, c)])
    while visit:
        rr, cc = visit.popleft()

        if rr < 0 or rr > n - 1 or cc < 0 or cc > m - 1 or arr[rr][cc] == "#":
            continue
        if arr[rr][cc] == "B":
            count = float("inf")
            break
        if visited[rr][cc]:
            continue
        if arr[rr][cc] == "G":
            count -= 1
        visited[rr][cc] = True
        visit.append((rr + 1, cc))
        visit.append((rr - 1, cc))
        visit.append((rr, cc + 1))
        visit.append((rr, cc - 1))


for _ in range(int(inp())):
    n, m = map_inp(int)
    arr = []
    for i in range(n):
        arr.append(list(inp()))

    visited = [[False for col in range(m)] for row in range(n)]
    count = 0
    for i in range(n):
        for j in range(m):
            if not visited[i][j] and arr[i][j] == "B":
                flood_fill(i, j, arr[i][j])
            if arr[i][j] == "G":
                count += 1
    bfs(n - 1, m - 1)
    if count == 0:
        print("YES")
    else:
        print("NO")