I always thought that my old friend John was rather richer than he looked, but I never knew exactly how much money he actually had. One day (as I was plying him with questions) he said:

* "Imagine I have between `m` and `n` Zloty..." (or did he say Quetzal? I can't remember!)
* "If I were to buy **9** cars costing `c` each, I'd only have 1 Zloty (or was it Meticals?) left."
* "And if I were to buy **7** boats at `b` each, I'd only have 2 Ringglets (or was it Zloty?) left."

Could you tell me in each possible case:

1. how much money `f` he could possibly have ?
2. the cost `c` of a car?
3. the cost `b` of a boat?

So, I will have a better idea about his fortune. Note that if `m-n` is big enough, you might have a lot of possible answers. 

Each answer should be given as `["M: f", "B: b", "C: c"]` and all the answers as `[ ["M: f", "B: b", "C: c"], ... ]`. "M" stands for money, "B" for boats, "C" for cars.

**Note:** `m, n, f, b, c` are positive integers, where `0 <= m <= n` or `m >= n >= 0`. `m` and `n` are inclusive.


## Examples:
```
howmuch(1, 100)      => [["M: 37", "B: 5", "C: 4"], ["M: 100", "B: 14", "C: 11"]]
howmuch(1000, 1100)  => [["M: 1045", "B: 149", "C: 116"]]
howmuch(10000, 9950) => [["M: 9991", "B: 1427", "C: 1110"]]
howmuch(0, 200)      => [["M: 37", "B: 5", "C: 4"], ["M: 100", "B: 14", "C: 11"], ["M: 163", "B: 23", "C: 18"]]
```

Explanation of the results for `howmuch(1, 100)`:

* In the first answer his possible fortune is **37**:
 * so he can buy 7 boats each worth 5: `37 - 7 * 5 = 2`
 * or he can buy 9 cars worth 4 each: `37 - 9 * 4 = 1`
* The second possible answer is **100**:
 * he can buy 7 boats each worth 14: `100 - 7 * 14 = 2`
 * or he can buy 9 cars worth 11: `100 - 9 * 11 = 1`

# Note
See "Sample Tests" to know the format of the return.

Difficulty: EASY
Tags: Fundamentals

Starter code:
def howmuch(m, n):
	

Solutions:
def howmuch(m, n):
	return [['M: %d' % i, 'B: %d' % (i / 7), 'C: %d' % (i / 9)] for i in range(min(m, n), max(m, n) + 1) if i % 7 == 2 and i % 9 == 1]


def howmuch(m, n):
	i = min(m, n)
	j = max(m, n)
	res = []
	while i <= j:
		if i % 9 == 1 and i % 7 == 2:
			res.append(['M: ' + str(i), 'B: ' + str(i // 7), 'C: ' + str(i // 9)])
		i += 1
	return res


def howmuch(m, n):
	(m, n) = sorted([m, n])
	return [['M: {}'.format(x), 'B: {}'.format(x // 7), 'C: {}'.format(x // 9)] for x in range(-((m - 37) // -63) * 63 + 37, n + 1, 63)]


from math import ceil

def howmuch(m, n):
	(m, n) = (min(m, n), max(m, n))
	x = max(0, ceil((m - 37) / 63))
	return [[f'M: {y}', f'B: {(y - 2) // 7}', f'C: {(y - 1) // 9}'] for y in range(37 + 63 * x, n + 1, 63)]


howmuch = lambda m, n: [[f'M: {M}', f'B: {M // 7}', f'C: {M // 9}'] for M in range(min(m, n), max(m, n) + 1) if M % 7 == 2 and M % 9 == 1]


def howmuch(m, n):
	val = []
	if m > n:
		(m, n) = (n, m)
	for i in range(m, n + 1):
		if (i - 2) % 7 == 0 and (i - 1) % 9 == 0:
			val.append(['M: {}'.format(i), 'B: {}'.format(int((i - 2) / 7)), 'C: {}'.format(int((i - 1) / 9))])
	return val


def howmuch(m, n):
	parameters = [m, n]
	parameters.sort()
	trueInstances = []
	for instance in range(parameters[0], parameters[1] + 1):
		if (instance - 1) % 9 == 0 and (instance - 2) % 7 == 0:
			carPrice = int((instance - 1) / 9)
			boatPrice = int((instance - 2) / 7)
			statement = [f'M: {instance}', f'B: {boatPrice}', f'C: {carPrice}']
			trueInstances.append(statement)
	return trueInstances


def howmuch(m, n):
	arr = []
	for i in range(min(m, n), max(m, n) + 1):
		if ((i - 2) / 7).is_integer() and ((i - 1) / 9).is_integer():
			x = int((i - 2) / 7)
			y = int((i - 1) / 9)
			arr.append(['M: {0}'.format(i), 'B: {0}'.format(x), 'C: {0}'.format(y)])
	return arr