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: Time Limit Exceeded)

#include<iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<map>
#include<queue>
#include<cmath>
#include <numeric>
#include<iomanip>
#include <cstdint>
#include<deque>
#include<stdio.h>
typedef long long ll;

using namespace std;


ll gcd(ll x, ll y) {
	if (y == 0) {
		return x;
	}
	else {
		return gcd(y, x % y);
	}
}


ll ged(ll num) {
	ll dig = 0;
	while (num != 0) {
		num /= 10;
		dig++;
	}
	return dig;
}

int main() {
	ll n;
	cin >> n;
	ll ans = 0;
	if (n % 2 == 1) {
		cout << 0 << endl;
	}
	else {
		ll rui = 1;
		while (rui <= n) {
			for (int i = 1; i < 1; i++) {
				for (int j = 0; j < i; j++) {
					rui *= 10;
				}
				cout << rui << endl;
				ans += n / rui;
			}
		}
		cout << ans << endl;
	}
}