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;
typedef long long ll;
const ll INF= 1LL << 60;
#define REP(i,n) for(ll i=0; i<n; i++)
#define FOR(i,a,n) for(ll i=a; i<n; i++)
struct Point{
    double x,y;
};
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);  
    ll n;
    cin >> n;
    ll cnt= 0;
    ll p=5;
    n-=n%5;
    while(n>p){
        p*=5;
        cnt++;
    }
    ll c= 0;
    FOR(i,1,cnt+1){
        c+= n/powl(5,i)/2;
    }
    
        

        
    cout << c << endl;
    

}