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;

// 問題文の式を提示する関数
// 動的計画法でなんとか高速化できないだろうか
// TODO: 動的計画法を勉強する
long long b(long long _N) {
  if(_N < 2) {
    return 1;
  }
  
  return _N * b(_N - 2);
}

int main(){
  long long N;
  cin >> N;
  long long zero = 0;
  
  // 計算するパート
  string num = to_string(b(N));
  
  // 末尾の0の個数を算出するパート
  for(int i = num.size(); i > 0; i--) {
    if(num.at(i - 1) == '0') {
      zero++;
    } else {
      break;
    }
  }
  
  cout << zero <<endl;
}