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;
#define INF 1<<30
#define endl '\n'
#define maxn 1000005
#define FASTIO ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
typedef long long ll;
const double PI = acos(-1.0);
const ll mod = 1e9 + 7;

inline void normal(ll &a) { a %= mod; (a < 0) && (a += mod); }
inline ll modMul(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); return (a * b) % mod; }
inline ll modAdd(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); return (a + b) % mod; }
inline ll modSub(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); a -= b; normal(a); return a; }
inline ll modPow(ll b, ll p) { ll r = 1; while (p) { if (p & 1) r = modMul(r, b); b = modMul(b, b); p >>= 1; } return r; }
inline ll modInverse(ll a) { return modPow(a, mod - 2); }
inline ll modDiv(ll a, ll b) { return modMul(a, modInverse(b)); }


///**
template < typename F, typename S >
ostream& operator << ( ostream& os, const pair< F, S > & p ) {
    return os << "(" << p.first << ", " << p.second << ")";
}

template < typename T >
ostream &operator << ( ostream & os, const vector< T > &v ) {
    os << "{";
    for (auto it = v.begin(); it != v.end(); ++it) {
        if ( it != v.begin() ) os << ", ";
        os << *it;
    }
    return os << "}";
}

template < typename T >
ostream &operator << ( ostream & os, const set< T > &v ) {
    os << "[";
    for (auto it = v.begin(); it != v.end(); ++it) {
        if ( it != v.begin()) os << ", ";
        os << *it;
    }
    return os << "]";
}

template < typename F, typename S >
ostream &operator << ( ostream & os, const map< F, S > &v ) {
    os << "[";
    for (auto it = v.begin(); it != v.end(); ++it) {
        if ( it != v.begin() ) os << ", ";
        os << it -> first << " = " << it -> second ;
    }
    return os << "]";
}

#define dbg(args...) do {cerr << #args << " : "; faltu(args); } while(0)

clock_t tStart = clock();
#define timeStamp dbg("Execution Time: ", (double)(clock() - tStart)/CLOCKS_PER_SEC)

void faltu () { cerr << endl; }

template <typename T>
void faltu( T a[], int n ) {
    for (int i = 0; i < n; ++i) cerr << a[i] << ' ';
    cerr << endl;
}

template <typename T, typename ... hello>
void faltu( T arg, const hello &... rest) { cerr << arg << ' '; faltu(rest...); }
// Program showing a policy-based data structure.
#include <ext/pb_ds/assoc_container.hpp> // Common file 
#include <ext/pb_ds/tree_policy.hpp>
#include <functional> // for less 
#include <iostream>
using namespace __gnu_pbds;
using namespace std;
// GNU link : https://goo.gl/WVDL6g
typedef tree<int, null_type, less_equal<int>, rb_tree_tag,
        tree_order_statistics_node_update>
        new_data_set;

// find_by_order(k) – ফাংশনটি kth ordered element এর একটা পয়েন্টার রিটার্ন করে। অর্থাৎ তুমি চাইলেই kth ইন্ডেক্সে কি আছে, সেটা জেনে ফেলতে পারছো!
// order_of_key(x) – ফাংশনটি x এলিমেন্টটা কোন পজিশনে আছে সেটা বলে দেয়।

/**___________________________________________________**/


ll trailingZero(ll n) 
{ 
    // Initialize result 
    ll count = 0; 
  
    // Keep dividing n by powers of  
    // 5 and update count 
    for (ll i = 5; n / i >= 1; i *= 5) 
        count += n / i; 
  
    return count; 
} 
int main()
{
    FASTIO
    /*
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    freopen("error.txt", "w", stderr);
#endif
//*/
    ll n;
    cin >> n;
    cout << trailingZero(n/2ll + 1ll) << endl;
}