Given an array of integers `a` and integers `t` and `x`, count how many elements in the array you can make equal to `t` by **increasing** / **decreasing** it by `x` (or doing nothing).
*EASY!*

```python
# ex 1

a = [11, 5, 3]
t = 7
x = 2

count(a, t, x) # => 3
```
- you can make 11 equal to 7 by subtracting 2 twice
- you can make 5 equal to 7 by adding 2
- you can make 3 equal to 7 by adding 2 twice

```python
# ex 2

a = [-4,6,8]
t = -7
x = -3

count(a, t, x) # => 2
```

## Constraints
**-10^(18) <= a[i],t,x <= 10^(18)**

**3 <= |a| <= 10^(4)**

Source: codewars | Difficulty: EASY
Tags: Fundamentals

Starter code: def count(a, t, x):
	

Solutions:
def count(a, t, x):
	return sum((not (t - v) % x if x else t == v for v in a))


count = lambda a, t, x: sum((x and (not (e - t) % x) or e - t == 0 for e in a))


def count(a, t, x):
	return sum(((t - i) % x == 0 for i in a)) if x else sum((i == t for i in a))


def count(a, t, x):
	if x == 0:
		return a.count(t)
	return sum(((t - e) % x == 0 for e in a))


def count(A, t, x):
	return sum((a % x == t % x for a in A)) if x else sum((a == t for a in A))


def count(lst, target, mod):
	if mod == 0:
		return lst.count(target)
	return sum(((n - target) % mod == 0 for n in lst))


def count(a, t, x):
	return sum((1 for n in a if (n % x == t % x if x else n == t)))


def count(a, t, x):
	result = 0
	for n in a:
		if n == t or (x != 0 and (n - t) % x == 0):
			result += 1
	return result


count = lambda a, t, x: sum(((e - t) % x == 0 if x != 0 else e - t == 0 for e in a))