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)

//#define _CRT_SECURE_NO_WARNINGS
#include "bits/stdc++.h"

#define rep(i,n) for(ll (i)=0;(i)<(ll)(n);(i)++)
#define all(x) (x).begin(),(x).end()
#define MOD 1000000007LL
#define INF (1LL<<60LL)
#define int long long
typedef long long ll;

using namespace std;


signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	ll n; cin >> n;
	if (n % 2 == 1)cout << 0 << endl;
	else {
		ll ans = 0;
		ll k = 10;
		while (n >= k) {
			ans += n / k;
			k *= 5;
		}
		cout << ans << endl;
	}
}