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: Wrong Answer)

// ABEY SAALE
// QUESTION KHUD KR SOLUTION KYU DEKH RHA H
 
 
 
 
 
 
 
 
//MAAF KIJIYEGA ME GUSSE ME IDHAR UDHAR NIKAL JATA HU
#include<bits/stdc++.h>
#define mod 1000000007
#define fr first
#define se second
#define ll long long
#define PI 3.1415926535 
#define pb push_back
#define mpr make_pair
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define senky_bansal ios_base::sync_with_stdio(false);
#define IIIT_ALLAHABAD    cin.tie(NULL);cout.tie(NULL);
using namespace std;

ll fac[2000005];
void pre(){ fac[0]=1; fac[1]=1; for(int i=2;i<200005;i++) fac[i]=(i*fac[i-1])%mod;}
ll power(ll a,ll b){ll res=1;while(b){if(b&1) res=(res*a)%mod;a=(a*a)%mod;b=b/2;}return res%mod;}
ll ncr(ll n,ll r){return ((fac[n]*power(fac[r],mod-2))%mod*power(fac[n-r],mod-2))%mod;}
// DEKHIYE JI  WALID JO THE ABBA HAMARE WO YEH function CHHOD KR GYE THE

signed main()
{   



      
      senky_bansal
      IIIT_ALLAHABAD
      
      // pre();
      ll t;
      // cin>>t;
      t=1;
      while(t--)
      { 
        ll n;
        cin>>n;
        if(n==0 || n==1)
        cout<<"1"<<endl;
        else if(n%2==0)
        {
            ll p1 = n/2;
            ll ans =0;
            ll r1 = 5;
            ll i=5;
            while(p1>=i)
            {
                ans+=p1/i;
                if(p1/5>=i)
                i*=5;
                else
                break;
            }
            
        cout<<ans;
        }
        else
        {
            cout<<"0";
        }
         

      }
      






}