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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
int main()
{
    /*#ifndef ONLINE_JUDGE
    freopen("input.txt","r",stdin);
    freopen("outp.txt","w",stdout);
    #endif*/
        ll t,n,m,i,a,b,c,count=0;
        cin>>n;
        a=1;b=1;
        if(n<2)
            cout<<count;
        else
        {
            if(n%2==1)
                cout<<0;
            else
            {
            n=n-n%10;
            for(i=10;i<=pow(10,18);i=i*10)
            {
                count=count+n/i;
            }
            cout<<count;
            }
        }
}