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 <cmath>
#include <vector>
#include <string>
#include <algorithm>

using namespace std;

#define INF 1e9
#define MAXN 100005
#define MAXM 100005
#define MOD 1000000007
#define ll long long
#define vi vector<int>
#define vll vector<long long>
#define rep(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define pii pair<int, int>

ll cnt = 0;
ll N;

void solve(){
    if(N % 2 == 1){
        cout << 0 << endl;
        return;
    }

    N /= 2;

    while(N > 0){
        cnt += N / 5;
        N/= 5;
    }
    
    cout << cnt << endl;
}

int main(){
    cin >> N;

    solve();
}