Determine the number of trailing zeros in n! (n factorial). A trailing zero is produced by a factor of 10 = 2*5, and since factors of 5 are the scarcer, the count equals the number of times 5 divides n!.
Input format
One line: a non-negative integer n.
Output format
One line: the number of trailing zeros in n!.
Constraints
- 0 <= n <= 1000000000