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!

Source: codewars | Difficulty: EASY
Tags: Fundamentals

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)


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):
	ans = 0
	if n and m > 0:
		ans = n * m
	if ans <= 0:
		ans = 0
	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):
	return all([n > 0, m > 0]) and n * m


from operator import mul
from functools import reduce

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


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)))


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):
	if n | m > 0:
		p = n * m
		return p
	else:
		return 0


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


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


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


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


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


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):
	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):
	c = 0
	g = 0
	c = n * m
	if n < 0 or m < 0:
		return g
	else:
		return c


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, 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):
	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:
		return 0
	elif m < 0:
		return 0
	else:
		return c


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


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


def paperwork(n, m):
	if n <= 0:
		return 0
	elif m <= 0:
		return 0
	else:
		return n * m
paperwork(5, 3)