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)

#pragma GCC optimize("O3")
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pi;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<pi> vpi;
typedef vector<vi> vvi;
const int inf = 0x3f3f3f3f;
const ll linf = 123456789012345678;
const ll mod = 1000000007;
#define all(x) x.begin(), x.end()
#define debug(x) cerr << #x << " = " << x << endl
#define sz(x) ((int)(x).size())


int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);

    ll n;
    cin >> n;

    if(n%2 == 1){
        cout << 0 << endl;
        return 0;
    }

    ll ans = 0;
    for(ll d = 10; d <= 1000000000000000000; d*=5)
        ans += n/d;
    cout << ans << endl;
}