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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
#define ll long int

int main(){
    long int n;
    cin >> n;

    long int k = 1,c2 = 0,c5 = 0,v,l;

    if(n % 2 == 0) v = 2;
    else v = 3;

    long int y = n;

    if(n % 2 == 1){
        cout << 0 << endl;
        return 0;
    }

    for(long int i = 0; i <= y / 2; i += 2){
        k = n - i;
        l = v + i;
        while(k % 2 == 0 || k % 5 == 0 || l % 2 == 0 || l % 5 == 0){
                if(k % 2 == 0){c2++;k /= 2;}
                if(k % 5 == 0){c5++;k /= 5;}
                if(k > l && l % 2 == 0){c2++;l /= 2;}
                if(k > l && l % 5 == 0){c5++;l /= 5;}
        }
    }

    long int ans = min(c2,c5);

    cout << ans << endl;
}