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;
using Graph = vector<vector<int>>;

typedef long long ll;
typedef long double ld;

#define rep(i, n) for(long long i = 0; i < (n); i++)
#define sqrt(d) pow((long double) (d), 0.50)

const ll INF = 2e9;
const ll large_P = 1e9 + 7;




signed main() {
	ll N; cin >> N;
	ll res = 0LL;


	if(N%2 == 0 && N >= 10){
		int i = 1;
		ll digit = pow(10, i);

		while(pow(10, i) <= N){
			res += (ll) N / digit;
			i++; digit = pow(10, i);
		}
	}



	cout << res << endl;

}