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 int ll;
const ll mod = 1e9+7;

//繰り返し二乗法
ll EX(ll x,ll n){
    if(n == 0)return 1;
    else if(n%2 == 1)return EX(x,n-1)*x;
    else return EX(x*x,n/2);
};

ll solve(ll n,ll num){
    ll res = 0;
    ll now = 1;
    if(n == 2) now--;
    while(n/(2*EX(num,now)) > 0){
        res += n/(2*EX(num,now));
        now++;
    }
    return res;
}

int main(){
    ll n;
    cin >> n;
    if(n%2 == 1){
        cout << 0 << endl;
        return 0;
    }    
    //cout << solve(n,2) << " " << solve(n,5) << endl;
    cout << solve(n,5) << endl;
    return 0;
}