Buggy python submission (verdict: Runtime Error):
import math
import sys
from itertools import accumulate
from functools import reduce
import queue
from fractions import gcd

def lcm(a, b):
    return a * b // gcd(a, b)

def combination_count(n, r):
    return math.factorial(n) // (math.factorial(n - r) * math.factorial(r))

def permutations_count(n, r):
    return math.factorial(n) // math.factorial(n - r)

sys.setrecursionlimit(10000)

mod = 1000000007

"""
# 0埋め, 小数点出力桁指定時のときの出力
a = 100
b = 0.987654321
print("{0:06d}-{1:6f}".format(a,b))
000100-0.987654

# 文字列をリストに格納
char_list = list("abcd") # ["a","b","c","d"]

# 標準入力取得
## 文字列
    = sys.stdin.readline().rstrip()
    = list(sys.stdin.readline().rstrip())
    = [list(list(sys.stdin.readline().rstrip())) for i in range(h)] #文字列を分解して二次元配列入力

## 数値
    = int(sys.stdin.readline())
    = map(int, sys.stdin.readline().split())
    = list(map(int, sys.stdin.readline().split()))
    = [list(map(int,list(sys.stdin.readline().split()))) for i in range(h)]  # 二次元配列入力　二次元マップみたいな入力のとき
"""

if __name__ == "__main__":
    h,w = map(int, sys.stdin.readline().split())
    s =  [list(list(sys.stdin.readline().rstrip())) for i in range(h)]
    flag=0
    for i in range(h):
        for j in range(w): #２重for文で行列を一つずつ確認
            if s[i][j] == "#":
                if s[i-1][j] =="." and s[i+1][j] =="." and s[i][j-1] =="."  and s[i][j+1] ==".":#上下左右塗れなかった時
                    flag = 1
                    print("No")
                    exit()
    print("Yes" if flag==0 else "No")

Fixed submission:
H,W=map(int,input().split())
S=["."+input()+"." for i in range(H)]
S=["."*(W+2)]+S+["."*(W+2)]
flag=0
for i in range(H):
    for j in range(W):
        if S[i][j]=="#":
            if S[i-1][j]=="." and S[i+1][j]=="." and S[i][j-1]=="." and S[i][j+1]==".":
                flag=1
print("Yes" if flag==0 else "No")

Buggy python submission (verdict: Runtime Error):
import math
import sys
from itertools import accumulate
from functools import reduce
import queue
from fractions import gcd

def lcm(a, b):
    return a * b // gcd(a, b)

def combination_count(n, r):
    return math.factorial(n) // (math.factorial(n - r) * math.factorial(r))

def permutations_count(n, r):
    return math.factorial(n) // math.factorial(n - r)

sys.setrecursionlimit(10000)

mod = 1000000007

"""
# 0埋め, 小数点出力桁指定時のときの出力
a = 100
b = 0.987654321
print("{0:06d}-{1:6f}".format(a,b))
000100-0.987654

# 文字列をリストに格納
char_list = list("abcd") # ["a","b","c","d"]

# 標準入力取得
## 文字列
    = sys.stdin.readline().rstrip()
    = list(sys.stdin.readline().rstrip())
    = [list(list(sys.stdin.readline().rstrip())) for i in range(h)] #文字列を分解して二次元配列入力

## 数値
    = int(sys.stdin.readline())
    = map(int, sys.stdin.readline().split())
    = list(map(int, sys.stdin.readline().split()))
    = [list(map(int,list(sys.stdin.readline().split()))) for i in range(h)]  # 二次元配列入力　二次元マップみたいな入力のとき
"""

if __name__ == "__main__":
    h,w = map(int, sys.stdin.readline().split())
    s =  [list(list(sys.stdin.readline().rstrip())) for i in range(h)]
    flag=0
    for i in range(h):
        for j in range(w):
            if s[i][j] == "#":
                if s[i-1][j] =="." and s[i+1][j] =="." and s[i][j-1] =="."  and s[i][j+1] ==".":
                    flag = 1
    print("Yes" if flag==0 else "No")

Fixed submission:
H,W=map(int,input().split())
S=["."+input()+"." for i in range(H)]
S=["."*(W+2)]+S+["."*(W+2)]
flag=0
for i in range(H):
    for j in range(W):
        if S[i][j]=="#":
            if S[i-1][j]=="." and S[i+1][j]=="." and S[i][j-1]=="." and S[i][j+1]==".":
                flag=1
print("Yes" if flag==0 else "No")