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>
#define REP(i,n) for(int64_t i=0;i<(n);i++)
#define ALL(a)  (a).begin(),(a).end()
#define SORT(c) sort((c).begin(),(c).end())
#define REV(c) reverse((c).begin(),(c).end())
#define YES cout << "YES" << endl;
#define NO cout << "NO" << endl;
#define Yes cout << "Yes" << endl;
#define No cout << "No" << endl;
#define COUTDOUBLE(d,i) cout << setprecision((i)) << (d) << endl;
#define VIN(v) for(int64_t i=0;i<(v).size();i++){cin >> (v).at(i);}
#define MREP(i,m) for(auto i=(m).begin(); i != (m).end(); ++i) // auto key = i->first;auto value = i->second;

using namespace std;
typedef int64_t I;
typedef pair<I,I> P;
typedef vector<I> V;

I MOD = pow(10,9)+7;
I MAX = LONG_MAX;
I MIN = LONG_MIN;

// 最大公約数
I gcd(I a, I b) {
  return b != 0 ? gcd(b, a % b) : a;
}

// 最小公倍数
I lcm(I a, I b) {
  return a * b / gcd(a, b);
}

int ctoi(char c){
  return c - '0';
}

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  I n;
  cin >> n;
  if (n % 2 == 1){
    cout << 0 << endl;
  }else if (n ==0){
    cout << 1 << endl;
  }else{
    I t=0;
    REP(i,50){
      if (i > 0){
        I c = n / (pow(5,i) * 2);
        if (c > 0){
          t += c;
        }
      }
    }
    cout << t << endl;
  }
  return 0;
}