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 <iostream>
#include <stdio.h>
#include <string>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set> 
#include <map>
#include <queue>
#include <deque>
#include <stack>
#include <bitset>
#include <iomanip>
using namespace std;
#define assem99 ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define ll long long
#define ull unsigned long long
#define ld long double
#define f(i,a,b) for(int i=a; i<b; i++)
#define F first
#define S second
#define sz(a) (int)(a).size()
#define ReadFile freopen("input.txt","r",stdin)
#define WriteFile freopen("output.txt","w",stdout)
int x[] = { -1, -1, -1, 0, 0, 1, 1, 1 };
int y[] = { -1, 0, 1, -1, 1, -1, 0, -1 };
const int MOD = 1e9 + 7, N = 1e6 + 5;
const double pi = acos(-1);

int arr[N];

int main()
{
	assem99;

	ull ans = 0;
	ull n; cin >> n;
	if (n % 2 == 1) cout << '0';
	else
	{
		for (ull i = 10; i <= n; i *= 5)
		{
			ans += n / i;
		}
	}
	cout << ans;

	return 0;
}