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++ (GCC 9.2.1) (judge verdict: Accepted)


#pragma GCC optimize ("O3")

#include <bits/stdc++.h>

using namespace std;

#ifdef LOCAL
#include <prettyprint.hpp>
#define debug(...)  cerr << "[" << #__VA_ARGS__ << "]: ", d_err(__VA_ARGS__);
#else
#define debug(...) 83;
#endif

void d_err() {
    cerr << endl;
}

template <typename H, typename... T>
void d_err(H h, T... t) {
    cerr << h << " ";
    d_err(t...);
}

template <typename T>
void print(T x) {
    cout << x << endl;
}

#define ALL(x) (x).begin(), (x).end()
#define FOR(i, m, n) for (int i = (m); i < (n); ++i)
#define REVFOR(i, m, n) for (int i = (n - 1); i >= (m); --i)
#define REP(i, n) FOR(i, 0, n)
#define REVREP(i, n) REVFOR(i, 0, n)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define bcnt __builtin_popcountll

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef pair<ll,ll> Pll;
typedef pair<int,int> Pin;

ll INF = 1e16;
int inf = 1e9;
ll MOD = 1e9+7;

int main(){
    cin.tie(0);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(20);
    
    ll N;
    cin >> N;

    if (N&1) {
        print(0);
        return 0;
    }

    ll c2, c5;

    ll b = 2;
    while(N / b != 0) {
        c2 += N / b;
        b <<= 1;
    }

    b = 5;
    while(N / b != 0) {
        c5 += N / b / 2;
        b *= 5;
    }
    
    print(min(c2, c5));

}