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 <bits/stdc++.h>

using namespace std;

#define REP(var, n)  for (int var = 0; var < (n); var++)
#define RREP(var, n)  for (auto var = n - 1; var != static_cast<int>(-1); var--)
#define FOR(var, a, b)  for (auto var = (a); var < (b); var++)
#define RFOR(var, a, b)  for (auto var = b - 1; var != a; var--)

typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef pair<int, int> pii;

#define all(c) begin(c),end(c)

#define D(x) do { std::cerr << x << " "; } while (0)
#define D2(x) do { std::cerr << #x << ": " << x << std::endl; } while (0)


template<typename T>
std::ostream &operator<<(std::ostream &out, const std::vector<T> &v) {
    if (!v.empty()) {
        out << '[';
        std::copy(v.begin(), v.end(), std::ostream_iterator<T>(out, ", "));
        out << "\b\b]";
    }
    return out;
}


template<typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
    return os << '(' << p.first << ',' << p.second << ')';
}


template<class T>
bool chmax(T &a, const T &b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
bool chmin(T &a, const T &b) {
    if (b < a) {
        a = b;
        return true;
    }
    return false;
}

const int MOD = (int) 1e9 + 7;
const int INF = 1 << 29;

void solve() {
    ll n;
    cin >> n;
    ll res = 0;

    ll denom = 10;
    if (n % 2 == 0) {
        while(true) {
            ll t = (n / denom);
            if (t == 0) break;
            res += (n / denom);
            denom *= 5;
        }
    }

    cout << res << endl;
}


int main() {
    cout.precision(30);
    cin.tie(0);
    ios::sync_with_stdio(false);

//    cout << solve() << endl;
    solve();
//    cout << (solve() ? "Yes" : "No") << endl;
    return 0;
}