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 MOD 1000000007

int main() {

  long long N;
  cin >> N;

  if(N%2 != 0) cout << 0 << endl;
  else{
    long long n = N,ans = 0;
    vector<long long> v(20);
    v.at(0) = 1;
    for(int i = 0;i < 19;i++) v.at(i+1) = v.at(i)*10+1;

    for(int i = 18;i >= 1;i--) {
      long long b = 1;
      for(int j = 0;j < i;j++) b *= 10;
      while(n >= b) {
        ans += v.at(i-1);
        n -= b;
      }
    }
    cout << ans << endl;
  }

  
}