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)

#include<bits/stdc++.h>
#define lop(i,n,a) for(int i=0;i<n;i+=a)
#define rep(i,n,a) for(int i=a;i<n;i++)
#define inarr(arr,n) lop(i,n,1)cin>>arr[i]
#define ll long long int
#define pb push_back
#define all(v) v.begin(),v.end()
#define endl "\n"
#define eb emplace_back
#define fr first
#define sc second
#define pii pair<int,int>
#define IOS ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define ret return 0
#define pi 3.1415926536
#define hell 1000000007
using namespace std;
// syntax for min priority_queue of pairs
// priority_queue<pair<ll,ll>,vector<pair<ll,int>>,greater<pair<ll,int>>> q;
int solve(){
  ll n;cin>>n;
  if(n&1)cout<<0<<endl;
  else{
      ll i=10;
      ll sum=0;
      while(n>=i){
       sum=sum+n/i;
       i*=5;
      }
      cout<<sum<<endl;
  }
  ret;
}
int main(){
    IOS;
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    int t=1;//cin>>t;
    while(t--){
        solve();
    }
    ret;
}