Your classmates asked you to copy some paperwork for them. You know that there are 'n' classmates and the paperwork has 'm' pages.

Your task is to calculate how many blank pages do you need.

### Example:
```python
paperwork(5, 5) == 25
```

**Note:** if `n < 0` or `m < 0` return `0`!

Waiting for translations and Feedback! Thanks!

Difficulty: introductory

Starter code:

def paperwork(n, m):
	

Solutions:
def paperwork(n, m):
    return n * m if n > 0 and m > 0 else 0

def paperwork(n, m):
    if m<0 or n<0:
        return 0
    else:
        return n*m

def paperwork(n, m):
    return max(n, 0)*max(m, 0)

def paperwork(n, m):
    return n*m if n>=0 and m>=0 else 0

def paperwork(n, m):
    return 0 if n < 0 or m < 0 else n*m

def paperwork(n, m):
    if m > 0 and n > 0:
        return m*n
    else:
        return 0

paperwork = lambda a,b: a*b if min(a,b)>0 else 0

def paperwork(n, m):
    return n * m if min((n, m)) > 0 else 0

paperwork = lambda m,n: m*n if m>0 and n>0 else 0

def paperwork(n, m):
    return m*n * int(n > 0 and m > 0)

def paperwork(n, m):
    return n*m if n > 0 < m else 0

def paperwork(n,m): return m*n if (m>0) & (n>0) else 0

def paperwork(n, m):
  #Declaring the variable with a value makes it local instead of global
  ans = 0
  
  #Checks that n and m are > 0 before doing the math
  if n and m > 0:
    ans = n * m
  
  #Throws out any answers that are negative
  if ans <= 0:
    ans = 0
    
  #Returns the ans variable  
  return ans

def paperwork(n, m):
    if n >= 0 and m >= 0:
        return n * m
    elif n < 0 or m < 0:
       return 0

def paperwork(n, m):
    # Happy Coding! ^_^
    if n > 0 and m > 0:
        return n * m
    else:
        return 0

def paperwork(n, m):
    return 0 if min(n,m) < 0 else n*m

def paperwork(n, m):
    return n*m*(n>0)*(m>0)

def paperwork(n, m):
    return [0, n * m][n > 0 and m > 0]

def paperwork(n, m):
    return n * m * (n > 0 < m)

def paperwork(n, m):
    return all([n>0,m>0]) and n*m

def paperwork(n, m):
    # Happy Coding! ^_^
    if (n < 0) or (m < 0 ):
        return 0
    else : return n *m

from operator import mul
from functools import reduce

def paperwork(*args):
    return reduce(mul, args) if min(args) > 0 else 0

def paperwork(n, m):
    return n*m if n == abs(n) and m == abs(m) else 0

paperwork = lambda o_,_o: max(0, _o) * max(o_, 0)

paperwork = lambda n, m: max(max(n, 0)*m, 0)

paperwork = lambda n, m: 0 if(n < 0 or m < 0) else n*m #Another Stupid Comment By Mr. Stupido

paperwork = lambda n, m: n * m if n >= 0 and m >= 0 else 0


paperwork = lambda n, m: n * m if all(x > 0 for x in (n, m)) else 0

def paperwork(n, m):
    return sum(True for _ in range(n) for _ in range(m))

def paperwork(n: int, m: int) -> int:
    return max(0, n) * max(0, m)

paperwork=lambda n, m: __import__('functools').reduce(__import__('operator').mul, [max(q, 0) for q in [n, m]])

def paperwork(n, m):
    return n*m*(min(m,n)>=0)

def paperwork(n, m):
    return (0, m * n)[n > 0 and m > 0]

def paperwork(n, m):
    if m > 0:
        if n > 0:
            return m*n
        else:
            return 0
    else:
        return 0

def paperwork(n, m):
    if n | m > 0:
        p = n * m
        return p
    else:
        return 0

def paperwork(n, m):
    return n * m * (n>=0) * (m>=0)

paperwork = lambda n, m: n * m if n > 0 and m > 0 else 0

def paperwork(n: int, m: int):
    return n * m if n > 0 and m > 0 else 0

def paperwork(n, m):
    if n>0 and m>0:
        b = n * m
    else:
        b=0
    return b
    
    # Happy Coding! ^_^


def paperwork(n, m):
    
    if n*m<=0:
        return 0
    elif m<=0:
        return 0
    elif n<=0:
        return 0
    elif n*m > 0:
        BP = n*m
        return BP
    # Happy Coding! ^_^


def paperwork(n, m):
    z =  n*m
    if n <= 0:
        return 0
    elif m <= 0:
        return 0
    else:
        return z
    # Happy Coding! ^_^


def paperwork(n, m):
    result =0
    
    if n<0:
        result==0
    elif m<0:
        result ==0
    else:
        result=n*m
    
    return result

def paperwork(n=0, m=0):
    return 0 if n <= 0 or m <= 0 else n * m
    
    # Happy Coding! ^_^


def paperwork(n, m):
    if n < 0 or m < 0:
        return 0
    else:
        pages = (n * m)
        if pages < 0:
            return 0
        else:
            return pages

def paperwork(n, m):
    if n <= 0 or m <= 0:
        cevap = 0
    else:
        cevap = n*m
    return cevap

def paperwork(n, m):
    print(n,m)
    if n > 0 and m > 0:
        return n * m
    else:
        return 0

def paperwork(n, m):
    result = n * m
    if n > 0 and m > 0:
        result = n * m
    elif n < 0 or m < 0:
        result = 0
    return result

def paperwork(n, m):
    if (n < 0):
        return 0
    elif(m < 0):
        return 0
    elif(n or m > 0):
        return n*m



def paperwork(n, m):
    if n >0 and m > 0:
        return n*m
    elif n<0 or m<0 or n == 0 or m ==0:
        return 0
    # Happy Coding! ^_^


def paperwork(n,
              m):

    if (n < 0 or
        m < 0):

        return 0
        
    return n * m


def paperwork(n = 0, m= 0):
    if n < 0 or m < 0:
        return 0
    return m * n


def paperwork(n, m):
    if n > 0 and m > 0:
        pages = n * m
        return pages
    else:
        return 0

def paperwork(n, m):
    if (n <= 0) or (m <= 0):
        return 0
    elif (n > 0) and (m > 0):
        pages = n * m
        return pages

def paperwork(n, m):
    if n<0 or m<0:
        res = 0
    else:
        res = n*m
    return res

def paperwork(n, m):
    if n < 0 and m < 0 or n < 0 or m < 0:
        return 0
    
    else:
        return n*m 


def paperwork(n, m):
    # Happy Coding! ^_^
    # uwu
    if n > 0 and m > 0:
        return(n * m)
    else:
        return 0

def paperwork(n, m):
    if n < 0 or 0 > m:
        R = 0
    else:
        R = n * m
    return R

def paperwork(n, m):
    if n < 0 or m < 0:
        return 0
    c = n * m
    return c

def paperwork(n, m):
    if n < 0 or m < 0:
        return 0
    else:
        r = n * m
        return r 


paperwork=lambda x, y: max(0, x) * max(0, y)


def paperwork(n, m):
    # Happy Coding! ^_^
    blankPages = 0
    if n > 0 and m > 0:
        blankPages = n * m
    return blankPages
    


def paperwork(n, m):
    if n<0 or m<0:
        totpag=0
    else:
        totpag=n*m
            
    return totpag

def paperwork(n, m):
    if (n * m) <= 0:
        return 0
    if n <= 0:
        return 0 
    if m <= 0: 
        return 0
    else: 
         return (n * m)

def paperwork(n, m):
    
    while n > 0 and m > 0:
        
        return n * m
    
    else:
        
        return 0

def paperwork(n, m):
    
    if n < 0 or m < 0: return 0
    
    papers = n * m
    
    return papers

def paperwork(n, m):
    r = n * m
    return r if n > 0 and m > 0 else 0

def paperwork(n, m):
    total = n * m
    if n < 0 or m <0:
        return 0
    else:
        return total 

def paperwork(n, m):
    if n <= 0:
        return 0
    if m <= 0:
        return 0
    else:
        paper = n * m
        return paper

def paperwork(n, m):
    
    paperwork1 = n * m
    nothing = 0
    if n >= 0 and m >= 0:
        return paperwork1
    if n < 0 or m < 0:
        return nothing
    
    


paperwork=lambda n,m:0 if n+m!=abs(n)+abs(m) else n*m

def paperwork(n=5, m=5):
    if n + 0 < 0:
        return 0
    if m + 0 <0:
        return 0
    x = n*m
    if x < 0:
        return 0
    return x

def paperwork(n, m):
    n = n * m
    if n <= 0 or m < 0:
        return 0
    else:
        return n

def paperwork(n, m):
    c = 0
    g = 0
    c = n*m
    if n < 0 or m < 0:
        return g

    else:
        return c

def paperwork(n, m):
    if n<=0 or m<=0:
        return 0
    if n*m > 0:
        return n*m


def paperwork(n, m):
    if n <= 0:
        return 0
    if m <= 0:
        return 0
    else:
        result = n * m
        return result
    
   


#this function tells you how much paper you need based on the how many people their are and how many pages their are.
def paperwork(people, pages):
    blank=people*pages
    if people<0:
        blank=0
    if pages<0:
        blank=0
    return blank

def paperwork(n, m):
    return n * m if all([n >= 0 , m >= 0]) else 0

def paperwork(n: int, m: int):
    return 0 if n < 0 or m < 0 else n * m

def paperwork(n, m):
    if n<0 or m<0:
        return(0)
    else:
        pages = m*n
        return pages

def paperwork(n, m):
    if n<0 or m<0:
        return 0
    d=n*m
    return d
        
       
        
   
 
   
   
    
       
   
        


import operator

def paperwork(n, m):
    if (n<0 or m<0):
        return 0
    return operator.mul(n, m)

def paperwork(n, m):
    r = n * m
    
    if n < 0 or m < 0:
        return 0
    else:
        return r
    # Happy Coding! ^_^


def paperwork(n, m):
    if n>=0 and m>0:
        return n*m
    else:
        return 0


def paperwork(n: int, m: int) -> int:
    return 0 if n < 0 or m < 0 else n * m

def paperwork(n, m):
    # Happy Coding! ^_^
    if m>0 and n>0:
        papers=n*m
    else:
        papers=0
    return papers

def paperwork(n, m):
    work = n*m
    if n <0 or m <0:
        return 0 
    else:
        return work 

def paperwork(n, m):
    if n < 0 or m < 0:
        return 0;
    page_amount = n * m
    return page_amount

def paperwork(n, m):
    result = 0
    if ((m > 0) & (n > 0)):
        result = n * m
    return result

def paperwork(n, m):
    c = n*m
    if n<0: #checking the first arg being more than zero
        return 0
    elif m<0: #checking second arg being more than zero
        return 0
    else: #returning papers
        return c
    # Happy Coding! ^_^


def paperwork(n, m):
    if n < 0 or m < 0:
        return 0
    else:
        return n * m
    # Happy Coding! ^_^ :D (✿◠‿◠)


def paperwork(n, m):
    if m<0:
        return 0
    if n<0:
        return 0
    if m==0:
        return 0
    if n==0:
        return 0
    if n>0 and m>0:
        return n*m

def paperwork(n, m):
    x = n*m
    if n < 0:
        return(0)
    elif m <0:
        return(0)
    else:
        return (x)

def paperwork(n, m):
    bp = n * m
    if n < 0 or m < 0:
        return 0
    else:
        return bp

def paperwork(n, m):
        if m<0 or n<0 :
           n = 0
        return n*m

def paperwork(n, m):
    if n > 0 and m > 0:
        blank_pages = n * m
    else:
        blank_pages = 0
    return blank_pages

def paperwork(n, m):
    if n < 0 and m < 0:
        return 0
    return max(n * m, 0) 

def paperwork(n, m):
    paperwork = n * m
    if n < 0 or m < 0:
        return 0
    else:
        return paperwork

def paperwork(n, m):
    # Happy Coding! ^_^
    # ty uwu
    if n < 0 or m < 0:
        return 0
    else:
        return n * m

def paperwork(n, m):
    if n<= 0:
      return 0
    elif m<=0:
        return 0
    else:
        return n * m
        
paperwork(5, 3)
    # Happy Coding! ^_^


def paperwork(n, m):
    return n*m*(m>0 and n>0)