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 pii=pair<int,int>;
using ll=long long;

#define fi first
#define se second
#define pb push_back
#define rep(i,n) for(int i=0; i<(int)(n); i++)

ll pow(ll a, int b) {
	ll c = 1;
	while(b > 0) {
		c*=a;
		b--;
	}
	return c;
}
int main() {

	ll n;
	cin >> n;
	ll a = 2;
	ll ans = 0;
	while(a <= n) {
		a *= 5;
		ans += n / a;
	}
	cout << ans << endl;

  return 0;
}