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 <string>
#include <vector>
#include <algorithm>
#include <cmath>
#include <iomanip>
#include <map>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep1(i, n) for (int i = 1; i < (int)(n) + 1; i++)
#define all(vecto) vecto.begin(), vecto.end()
#define max3(a, b, c) max(a, max(b, c))
#define min3(a,b,c) min(a,min(b,c))
typedef long long ll;

void ans(bool qus){
    if (qus == true)
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
}

void Ans(bool qust){
    if (qust == true)
        cout << "YES" << endl;
    else
        cout << "NO" << endl;
}

////////////////---START---///////////////

//愚直に解くと計算量がエグくなる

/*ll func(ll s){
    if (s < 2){
        return 1;
    }
    else
        return s * func(s - 2);
}

int main(){
    ll n;
    cin >> n;
    int pro = func(n);
    int i = 1;
    while(pro / pow(10, i) == 0.0){
        pro = pro / pow(10, i);
        i++;
    }
    cout << i << endl;
}*/

int main(){
    ll n;
    cin >> n;
    if (n % 2 != 0){
        cout << 0 << endl;
    }
    else{
        ll count;
        n /= 2;
        while(n){
            count += n / 5;
            n /= 5;
        }
        cout << count << endl;
    }
}