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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
 
 
typedef long long ll;
typedef pair<int,int> ii;
typedef vector<int> vi;
typedef vector<ii> vii;
typedef long double ld;
 
typedef set<int>::iterator sit;
typedef map<int,int>::iterator mit;
typedef vector<int>::iterator vit;
 
const int INF = 1e9 + 7;
const int MOD = 1e9 + 7;
const int MAXN = 1e6 + 3;
 
#define _  %  MOD
#define __ %= MOD
 
#define      each(it,s)        for(auto it = s.begin(); it != s.end(); ++it)
#define      sortA(v)          sort(v.begin(), v.end())
#define      sortD(v)          sort(v.begin(), v.end(), greater<auto>())
#define      fill(a)           memset(a, 0, sizeof (a))
 
#define      swap(a,b)        {a=a+b;b=a-b;a=a-b;}
#define      rep(i, n)         for(ll i = 0; i < (n); ++i)
#define      repA(i, a, n)     for(ll i = a; i <= (n); ++i)
#define      repD(i, a, n)     for(ll i = a; i >= (n); --i)
#define watch(x) cout << (#x) << " is " << (x) <<endl
 
#define fi first
#define se second
#define mp make_pair
#define pb push_back
 
#define fbo find_by_order
#define ook order_of_key
 
ll gcd(ll a,ll b){if(b==0) return a; else return gcd(b,a%b);}
ll power(ll x,ll y){ll res=1; while(y>0){if(y&1) res=res*x;y=y>>1;x=x*x;}return res;} // modular exponent
 
int main()
{
    ios_base::sync_with_stdio(false); // don't use printf and scanf
    cin.tie(NULL); // cout<<fixed<<setprecision
	ll n;
	cin>>n;
	if(n%2==0)
	{
		n/=2;
		ll ans=0;
		ll val=5ll;
		while(n/val)
		{
			ans+=n/val;
			val*=5;
		}
		cout<<ans<<'\n';
	}	
	else
	{
		ll val5=0;
		ll val2=0;
		ll val=5;
		while(n/val)
		{
			val5+=n/val;
			val*=5;
		}
		val=2;
		while(n/val)
		{
			val2+=n/val;
			val*=2;
		}
		val2-=(n/2);
		n/=2;
		val=5;
		while(n/val)
		{
			val5-=(n/val);
			val*=5;
		}
		val=2;
		while(n/val)
		{
			val2-=(n/val);
			val*=2;
		}
		ll ans=min(val5,val2);
		cout<<ans<<'\n';
	}
	return 0;
}
// JUST ASK YOURSELF DID YOU GIVE YOUR BEST ? ISSE ZYADA
// KUCH KAR BHI NAHI SAKTE !! ENJOY AND GIVE YOUR BEST!!