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)**

Difficulty: introductory

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)

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

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