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 <bits/stdc++.h>
using namespace std;
using ll=long long;
using vin=vector<int>;
using vll=vector<long long>;
using vvin=vector<vector<int>>;
using vvll=vector<vector<long long>>;
using vstr=vector<string>;
using vvstr=vector<vector<string>>;
using vch=vector<char>;
using vvch=vector<vector<char>>;
using vbo=vector<bool>;
using vvbo=vector<vector<bool>>;
using vpii=vector<pair<int,int>>;
using pqsin=priority_queue<int,vector<int>,greater<int>>;
#define mp make_pair
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep2(i,s,n) for(int i=(s);i<(int)(n);i++)
#define all(v) v.begin(),v.end()
#define decp(n) cout<<fixed<<setprecision((int)n)
const ll inf=1e9+7;
const ll INF=1e18;

int main(){
    ll n;cin>>n;
    if(n%2)cout<<0<<endl;
    else{
        n/=2;
        ll ans=(ll)0;
        ll c=5;
        while(c<=n){
            ans+=n/c;
            c*=5;
        }
        cout<<ans<<endl;
    }
}