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 fi first
#define endl "\n"
#define se second
#define ls( s) (s&(-s))
#define ll long long
#define inf 0x3f3f3f3f
const ll N = 300030;
#define rep(i, begin, end) for (__typeof(end) i = (begin) - ((begin) > (end)); i != (end) - ((begin) > (end)); i += 1 - 2 * ((begin) > (end)))
const ll mod =    1e9+7;
int n;
int arr[N];
int main()
{ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
ll x;
cin>>x;
ll ans=0;
ll cnt=1;
if(x%2)
cout<<0;
else{
ll y=0;
while(x)
{

	x/=5;
	ans+=x;


	
}
cout<<ans;}
}