Given two integers `a` and `x`, return the minimum non-negative number to **add to** / **subtract from** `a` to make it a multiple of `x`.

```python
minimum(10, 6)  #= 2

10+2 = 12 which is a multiple of 6
```

## Note
- 0 is always a multiple of `x`

## Constraints

**1 <= a <= 10^(6)**

**1 <= x <= 10^(5)**

Source: codewars | Difficulty: EASY
Tags: Fundamentals

Starter code: def minimum(a, x):
	

Solutions:
def minimum(a, x):
	return min(a % x, -a % x)


def minimum(a, x):
	return abs(a - x * round(a / x))


def minimum(a, x):
	r = a % x
	return r and min(r, x - r)


def minimum(a, x):
	near = x
	while near < a:
		near += x
	plus = near - a
	minus = a - (near - x)
	return plus if plus < minus else minus


def minimum(a, b):
	return min(a % b, b - a % b)


def minimum(a, x):
	n = a % x
	m = (a // x + 1) * x - a
	return min(n, m)