Problem:
Score : 500 points

Problem Statement

For an integer n not less than 0, let us define f(n) as follows:

f(n) = 1 (if n < 2)

f(n) = n f(n-2) (if n \geq 2)

Given is an integer N. Find the number of trailing zeros in the decimal notation of f(N).

Constraints

0 \leq N \leq 10^{18}

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of trailing zeros in the decimal notation of f(N).

Sample Input 1

12

Sample Output 1

1

f(12) = 12 × 10 × 8 × 6 × 4 × 2 = 46080, which has one trailing zero.

Sample Input 2

5

Sample Output 2

0

f(5) = 5 × 3 × 1 = 15, which has no trailing zeros.

Sample Input 3

1000000000000000000

Sample Output 3

124999999999999995

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

long long solve(long long N, long long x)
{
    long long ans = 0;
    unsigned long long curr = x;

    while(N >= curr)
    {
        ans += N / curr;
        curr *= x;
    }

    return ans;
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(NULL), cout.tie(NULL);

    long long N;
    cin >> N;

    if(N % 2)
    {
        cout << 0 << "\n";
        return 0;
    }
 
    long long two = N / 2 + solve(N / 2, 2);
    long long five = solve(N / 2, 5);

    cout << min(two, five) << "\n";
    return 0;
}