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: Wrong Answer)

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cctype>
#include <climits>
#include <string>
#include <bitset>
using namespace std;
typedef long double ld;
typedef long long int ll;
typedef unsigned long long int ull;
typedef vector<int> vi;
typedef vector<char> vc;
typedef vector<string> vs;
typedef vector<double> vd;
typedef vector<ll> vll;
typedef vector<pair<int,int> > vpii;
typedef vector<pair<string,int>> vpsi;
typedef vector<vector<int> > vvi;
typedef vector<vector<char> > vvc;
typedef vector<vector<string> > vvs;
typedef vector<vector<ll> > vvll;
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define fin(ans) cout<<(ans)<<endl
#define STI(s) atoi(s.c_str())
#define mp(p,q) make_pair(p,q)
#define pb(n) push_back(n)
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define Sort(a) sort(a.begin(),a.end())
#define Rort(a) sort(a.rbegin(),a.rend())
template<class T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;}return 0;}
template<class T> inline bool chmin(T& a,T b){if(a>b){a=b;return 1;}return 0;}
const int P = 1000000007;
const int INF = INT_MAX;
const ll LLINF = 1LL<<60;

int main(void){ios::sync_with_stdio(false);cin.tie(0);
//////////////////////////////////////////////////////

ll n;
cin >> n;
if(n%2==1){
    fin(0);
    return 0;
}
string N;
N = to_string(n);
int keta;
ll ans=0;
keta = N.length();
rep(i,keta){
    string temp;
    temp = N[i];
    int temp2;
    temp2 = STI(temp);
    ans += temp2 * (keta-i-1);
}
fin(ans);

//////////////////////////////////////////////////////
return 0;}