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)**

Difficulty: introductory

Starter code:

def minimum(a, x):
	

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

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


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


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


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

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


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):
    return min(a%x, x-a%x)

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