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;
typedef long long ll;
typedef vector<long long> vll;
typedef vector<pair<int,int>> vp;
typedef pair<int,int> pii;
#define INF (int)(1e9)
#define MAXX 1.1529215e+18
#define inf 999999
#define EPS (1e-7)
#define rep(i,n) for(int i=0; i<(int)(n);i++)
#define REP(i,n) for(int i=1;i<=(int)(n);i++)
#define FOR(i,k,n) for(int i=(k);i<(int)(n);i++)
#define ALL(a)  a.begin(),a.end()
#define RALL(a) a.begin(),a.end(),greater<int>()
#define ROT(a) a.begin(),a.begin()+1,a.end()
#define RROT(a) a.begin(),a.end()-1,a.end()
#define PB push_back
#define MP make_pair
#define PI acos(-1.0)
#define sz(a) a.size()
const ll MOD = 1e9+7;
template<typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { if (a > b) { a = b; return 1; } return 0; }
template<typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { if (a < b) { a = b; return 1; } return 0; }
template<typename T> T gcd(T a, T b) { if (b == 0) return a; return gcd(b, a % b); }
//cout << 'H' << endl;
/*--------------------------------------------*/

int main(){
    //cout.precision(10);
    cin.tie(0);
    ios::sync_with_stdio(false);

    ll N;
    cin >> N;

    if(N%2){
        cout << 0 << endl;
    }else{
        // int digit = sz(to_string(N));  // 45435 -> 5
        // int tmp = digit - 1;
        ll ans = 0;

        ll M = (N/10) * 10;

        ll tmp = 5;

        while(tmp < M){
            ans += ((M / tmp) / 2);
            tmp *= 5;
        }

        
        cout << ans << endl;
    }

}