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>
using namespace std;
#define ll long long 
// memo  fixed  setprecision(20);
using vvll = vector<vector<ll>>;
ll mod =1000000007;
/*"itob" int to "N"base */           template<typename TypeInt> string itob(const TypeInt v, int base){static const char table[] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";string ret;static numeric_limits<TypeInt> t;TypeInt n = v;if (t.is_signed) {if (v < 0) n *= -1;}while (n >= base) {ret += table[n%base];n /= base;}ret += table[n];if (t.is_signed) {if (v < 0 && base == 10) ret += '-';}reverse(ret.begin(), ret.end());return ret;}
/*"chmin" a = MAX*/                  template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
/*"chmax" a = MIN*/                  template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
/*"ctoi"  char to int*/              int ctoi(char c) {return c-'0';}
/*"gcd"   MAX Euclidean */           ll gcd(ll a,ll b){if(b==0)return a; return gcd(b,a%b);}
/*"lcm"   MIN*/                      ll lcm(ll a,ll b){ll g = gcd(a,b);return a/g*b;}
/*"primecheck"If prime,return true.*/bool primecheck(ll n){if(n < 2) return false;else{for(int i = 2; i * i <= n; i++){if(n % i == 0) return false;}return true;}}
string  reverserange(string s,ll a,ll b){reverse(s.begin()+a-1,s.begin()+b); return s;}
ll g1(ll n,ll p){
    if(n==0)return 0;
    ll res =n/p;
    res+=g1(n/p,p);
    return res;
}
ll g2(ll n,ll p){
    if(n%2)return 0;
    ll res =0;
    if(p==2)res+=n/2;
    res = g1(n/2,p);
    return res;
}
int main(){
    ll n;
    cin >> n;
    ll ans = min(g2(n,5),g2(n,2));
    cout << ans << endl;
    


}