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 <utility>
#include <functional>
#include <numeric>
#include <list>
#include <set>
#include <map>
#include <algorithm>
#include <cmath>
#include <limits>
#include <iomanip>
#include <bitset>
#include <queue>
#include <unordered_set>
#include <unordered_map>

using namespace std;
typedef long long ll;
typedef pair<int,int> pii;

#define rep(i, n) for(int i = 0;i < n;i++)
const int INF = 1LL << 30;
const int MOD = 1000000007;

int main(){
    std::ios_base::sync_with_stdio(false);
    cin.tie(0);

    ll n; cin >> n;

    if(n%2==1){
        cout << 0 << endl;
    } else {
        ll cnt=0;
        n/=2;

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