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 <iostream>
#include <bits/stdc++.h>

#define ll long long
#define ull unsigned long long
using namespace std;

ull f(ull n)
{
    ull  cnt = 0;
    for (int i = 5; n>=i; i *= 5)
        cnt += (ull)n/(ull)i;
    return cnt;
}

int main()
{
       ull n;
       cin>>n;
       //ull ans = f(n);
       ull ans = 0 ;
        if(n%2!=0)
        {
            cout<<0<<endl;
        }
        else
        {
            ull cnt2 =0; ull cnt5 = 0 ;
            for(ull i = 2;  i<=n ; i*=2)
            {
                cnt2+=n/i;
            }
            for(ull i = 5;  i<=n ; i*=5)
            {
                cnt5+=(n/i)/2;
            }

            cout<<min(cnt2,cnt5)<<endl;
        }
        return 0;
}