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;

//conversion
//------------------------------------------
inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}

//math
//-------------------------------------------
template<class T> inline T sqr(T x) {return x*x;}

//typedef
//------------------------------------------
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<bool> VB;
typedef vector<double> VD;
typedef vector<string> VS;
typedef vector<char> VC;
typedef pair<int, int> PII;
typedef long long LL;
typedef vector<LL> VLL;
typedef vector<VLL> VVLL;

//container util
//------------------------------------------
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define MP make_pair
#define SZ(a) int((a).size())
#define SQ(a) ((a)*(a))
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) sort((c).begin(),(c).end())
#define LB lower_bound
#define UB upper_bound

//repetition
//------------------------------------------
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)

//constant
//--------------------------------------------
const double EPS = 1e-10;
const double PI  = acos(-1.0);
const LL INF = 1e9;
const LL MOD = 1e9+7;

//other
//-------------------------------------------
#define chmin(x,y) x=min(x,y)
#define chmax(x,y) x=max(x,y)
#define FI first
#define SE second

//memo
// setprecision()


LL Gcd(LL a, LL b) {
  if (a < b) {
    return Gcd(b,a);
  }
  if (b == 0) {
    return a;
  }
  else {
    return Gcd(b,a%b);
  }
}

LL Lcm(LL a, LL b) {
  LL abd = Gcd(a,b);
  LL x = a / abd, y = b / abd;
  
  return x*y*abd;
}

LL solve(LL n) {
  if(n%2){
    return 0;
  }
  n /= 2;
  LL ans=0;
  
  while(n > 0){
    ans += n/5;
    n /= 5;
  }
  return ans;
}

int main() {
  LL n;
  cin >> n;
  
  cout << solve(n) << endl;
  
  return 0;
  
}