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>
#include <stdio.h>

using namespace std;
#define inputInt(a) int a; cin >> a; 
#define inputInt2(a,b) int a; int b; cin >> a >> b; 
#define inputInt3(a,b,c) int a; int b; int c; cin >> a >> b >> c;
#define inputLong(a) long a;cin >> a; 
#define inputIntArray(a,N) int a[N];for(int i=0;i<N;i++){cin >> a[i];}
#define inputLongArray(a,N) long a[N];for(int i=0;i<N;i++){cin >> a[i];}
#define inputIntArray2(a,b,N) int a[N]; int b[N]; for(int i=0;i<N;i++){cin >> a[i] >> b[i];}
#define output(answer) cout << answer << endl;
#define fN(i,N)    for(int i=0; i<N; i++)
#define fSE(i,s,e) for(int i=s; i<=e; i++)
#define ll long long
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define FOR(i,N) for(int i=0; i<N; i++)
#define cinv(v,N) vector<int> v(N); for(int i=0; i<N; i++){ cin >> v[i]; }
#define mt make_tuple
const ll MOD2 = 998244353;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

ll power(ll x, ll y){//繰り返し2乗法
  if(y < 0){
    return 0;
  }else if(y == 0){
    return 1LL;
  }else if(y%2 == 0){//偶数
    return power(x,y/2)*power(x,y/2);
  }else{//奇数
    return power(x,y-1)*x;
  }
}

int main()
{
  //input
  cin.tie(0);
  ios::sync_with_stdio(false);
  
  ll N; cin >> N;
  
  //main
  if(N%2 == 1){ cout << 0 << endl; }
  else{
    ll ans = 0;
    for(ll i=1; i<=25; i++){
      ans += (N/2)/(power(5,i));
      //cout << ans << endl;
    }
    cout << ans << endl;
  }
  
  // output

  return 0;
}