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)

/*....Abhilash's coding template....*/
#include<bits/stdc++.h>
using namespace std;
#define endl "\n"
#define ll long long
#define pb push_back
#define pob pop_back
#define mp make_pair
#define in insert
#define ff first
#define ss second
#define all(a) (a).begin(),(a).end()
#define vll vector<ll>
#define fl(i,a,b) for(ll i=a;i<=b;i++)
#define flr(i,a,b) for(ll i=a;i>=b;i--)
#define print(v) for(auto i=v.begin();i!=v.end();i++){cout<<*i<<" ";}cout<<"\n";
#define tr(c,i) for(decltype((c).begin()) i = (c).begin(); i != (c).end(); i++)
#define present(c,x) ((c).find(x) != (c).end())
#define cpresent(c,x) (find(all(c),x) != (c).end())
#define test() ll te; cin >> te; while(te--)
#define DEBUG(x) cout << '>' << #x << ':' << x << endl;
#define PI 3.14159265
#define MOD 1000000007
#define M 998244353
#define node pair<ll , pair<ll,ll> >
#define fast ios_base::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL);

int main()
{
    ll ans=0,n; cin>>n;
    if(n&1)cout<<0;
    else{n/=2;
    while(n!=0)
    {
        ans+=(n/5);
        n/=5;
    }
    cout<<ans;
    }
}