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: Runtime Error)

#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define PI acos(-1)
#define pi 3.1415926535897932384
#define INF 2147483647
#define EPS 1e-8
#define MOD 1000000007
#define MAX 300005
using namespace std;

typedef long long ll;
typedef pair<int, int> ii;

int Set(int mask, int pos){return mask = mask | (1<<pos);}
bool check(int mask, int pos){return (bool)(mask & (1<<pos));}

int main()
{
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    #ifndef ONLINE_JUDGE
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout);
    #endif
    ll n;

    cin>>n;

    ll p = 10, cnt = 1;
    ll sum = 0;
    while(p <= n){
        sum += (n / p);
        p *= 5;
    }

    cout<<sum<<endl;

    return 0;
}