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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define ll long long
ll findTrailingZeros(ll n)
{
    // Initialize result
    ll count = 0;

    // Keep dividing n by powers of
    // 5 and update count
    for (ll i = 5; n / i >= 1; i *= 5)
        count += n / i;

    return count;
}
int main()
{
    ios_base::sync_with_stdio(0); cin.tie(0);
    ll x;
    cin>>x;
    if(x<2) return cout<<1, 0;
    ll ans = findTrailingZeros(x);
    if(x%5==0)
    {
        cout<< max(0ll,ans/2 - 2);
    }
    else cout<<ans/2;
}