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 rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep2(i,a,b) for(int i=a;i<b;i++)
#define each(it,v) for(auto it : v)
#define mod 1000000007
#define All(v) (v).begin(),(v).end()
#define ll long long
#define vi vector<int>
#define vll vector<ll>
#define vpii vector<pair<int,int>>
#define vv vector<vector<int>>
template<class T> T gcd(T a,T b) {return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b) {return a/gcd(a,b)*b; }
template<class T> bool isPrime(T num)
{
    if (num < 2) return false;
    else if (num == 2) return true;
    else if (num % 2 == 0) return false;

    double sqrtNum = sqrt(num);
    for (T i = 3; i <= sqrtNum; i += 2)
    {
        if (num % i == 0)
        {
            return false;
        }
    }
    return true;
}

using namespace std;

main()
{
    ll n;
    cin>>n;

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

    ll ans=0,tmp=5;
    while(n>tmp)
    {
        ans+=n/(tmp*2);
        tmp*=5;
    }
    cout<<ans<<endl;
}