Write a program that will calculate the number of trailing zeros in a factorial of a given number.
`N! = 1 * 2 * 3 *  ... * N`
Be careful `1000!` has 2568 digits...
For more info, see: http://mathworld.wolfram.com/Factorial.html 
## Examples
```python
zeros(6) = 1
# 6! = 1 * 2 * 3 * 4 * 5 * 6 = 720 --> 1 trailing zero
zeros(12) = 2
# 12! = 479001600 --> 2 trailing zeros
```
*Hint: You're not meant to calculate the factorial. Find another way to find the number of zeros.*
def zeros(n): 
      
    # Initialize result 
    count = 0
  
    # Keep dividing n by 
    # powers of 5 and 
    # update Count 
    i=5
    while (n/i>=1): 
        count += int(n/i) 
        i *= 5
  
    return int(count) 
def zeros(n): 
      
    # Initialize result 
    count = 0
  
    # Keep dividing n by 
    # powers of 5 and 
    # update Count 
    i=5
    while (n/i>=1): 
        count += int(n/i) 
        i *= 5
  
    return int(count) 