Let f(A) be the number of zeroes at the end of A!. (Recall that x! = 1 * 2 * 3 * … * x, and by convention, 0! = 1.)
For example, f(3) = 0 because 3! = 6 has no zeroes at the end, while f(11) = 2 because 11! = 39916800 has 2 zeroes at the end.
Given A, find how many non-negative integers x have the property that f(x) = A.
First and only argument to input is a single integer A.
Return a single integer denoting number of non-negative integers x having f(x) = A.
0 <= A <= 10^9
Input 1: A = 0 Output 1: 5 Explanation: 0!, 1!, 2!, 3!, and 4! end with K = 0 zeroes. Input 2: A = 5 Output 2: 0
NOTE: You only need to implement the given function. Do not read input, instead use the arguments to the function. Do not print the output, instead return values as specified. Still have a doubt? Checkout Sample Codes for more details.