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 <iostream> // cout, endl, cin
#include <string> // string, to_string, stoi
#include <vector> // vector
#include <algorithm> // min, max, swap, sort, reverse, lower_bound, upper_bound
#include <utility> // pair, make_pair
#include <tuple> // tuple, make_tuple
#include <cstdint> // int64_t, int*_t
#include <cstdio> // printf
#include <map> // map
#include <queue> // queue, priority_queue
#include <set> // set
#include <stack> // stack
#include <deque> // deque
#include <unordered_map> // unordered_map
#include <unordered_set> // unordered_set
#include <bitset> // bitset
#include <cmath>

using namespace std;
#define ll long long
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)



int main() {
	ll N;
	cin >> N;

	ll sum = 0;
	if (N % 2 == 0) {
		for (ll i = 5; i <= N; i = i * 5) {
			sum += N / (i * 2LL);
		}
	}
	cout << sum << endl;
	

}