In this Kata, you will be given two positive integers `a` and `b` and your task will be to apply the following operations:

```
i) If a = 0 or b = 0, return [a,b]. Otherwise, go to step (ii);
ii) If a ≥ 2*b, set a = a - 2*b, and repeat step (i). Otherwise, go to step (iii);
iii) If b ≥ 2*a, set b = b - 2*a, and repeat step (i). Otherwise, return [a,b].
```

`a` and `b` will both be lower than 10E8.

More examples in tests cases. Good luck!

Please also try [Simple time difference](https://www.codewars.com/kata/5b76a34ff71e5de9db0000f2)

Difficulty: EASY
Starter code: def solve(a,b):
	

Solutions:
def solve(a, b):
	if not (a and b):
		return [a, b]
	if a >= 2 * b:
		return solve(a % (2 * b), b)
	if b >= 2 * a:
		return solve(a, b % (2 * a))
	return [a, b]


def solve(a, b):
	if a == 0 or b == 0:
		return [a, b]
	elif a >= 2 * b:
		return solve(a % (2 * b), b)
	elif b >= 2 * a:
		return solve(a, b % (2 * a))
	else:
		return [a, b]


def solve(a, b):
	while a and b:
		if a >= b * 2:
			a %= 2 * b
		elif b >= a * 2:
			b %= 2 * a
		else:
			break
	return [a, b]


def solve(a, b):
	return [a, b] if not a or not b else solve(a % (2 * b), b) if a >= 2 * b else solve(a, b % (2 * a)) if b >= 2 * a else [a, b]


def solve(a, b):
	return [a, b] if a * b == 0 or 0.5 < a / b < 2 else solve(a % (2 * b), b % (2 * a))


def solve(a, b):
	for _ in range(0, 25):
		if b:
			a %= 2 * b
		if a:
			b %= 2 * a
	return [a, b]


def solve(a, b):
	while a and b and (a >= 2 * b or b >= 2 * a):
		if b:
			a %= 2 * b
		if a:
			b %= 2 * a
	return [a, b]