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: Accepted)

#include<bits/stdc++.h>
using namespace std;

#define ll long long
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define endl "\n"
#define pb push_back
#define all(a) (a).begin(),(a).end()
#define rev(a) (a).rbegin(),(a).rend()
#define pll pair<ll,ll>
#define rep(i,a,b) for(long long i = (a); i <= (b); i++)

const int mod = 1e9+7;
const int N = 1e5+5; 

int main(){
	fast;
	ll n; cin >> n;
	ll ans = 0;
	
	if(n & 1){
	    cout << "0\n";
	    
	}
	else{
	
	ll d = 10;
	while(n >= d){
	    ans += n / d;
	    d *= 5;
	}
	
	cout << ans << endl;
	}
}