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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdio>
#include <iomanip>
#include <fstream>
#include <cassert>
#include <cstring>
#include <numeric>
#include <ctime>
#include <complex>
#include <bitset>
#include <random>
#include <climits>
#include <stack>

using namespace std;

typedef long long ll;
typedef long double ld;

#define int ll
#define double ld
#define loop(i, n) for(int i = 0; i < (int)n; ++i)
#define loop1(i, n) for(int i = 1; i <= (int)n; ++i)
#define F first
#define S second
#define pb push_back
#define pi pair <int, int>
#define all(x) begin(x), end(x)
#define ti tuple <int, int, int>
#define Point Vect
#define no {cout << 0; return;}
#define mkp make_pair
#define mkt make_tuple
#define cerr if(0) cerr

void solve() {
    int n; cin >> n;
    if (n % 2) {
        cout << 0; return;
    }
    n >>= 1;
    int ans = 0;
    while (n) {
        n /= 5; 
        ans += n;
    }
    cout << ans;
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    solve();
    return 0;
}