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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
using pint = pair<int,int>;
ll mod = 1000000007,mod2 = 998244353;

void chmin(long long &a, long long b) { if (a > b) a = b; }
void chmax(long long &a, long long b) { if (a < b) a = b; }
const long long INF = 1LL<<60;
ll gcd(ll a,ll b){
    if(b == 0) return a;
    else return gcd(b,a%b);
}

ll lcm(ll a, ll b){
    ll g = gcd(a, b);
    return (a / g) * b;
}

//重みある時
using Edge = pair<int,int>;
using Graph = vector<vector<Edge> >;

ll cnt2 = 0,cnt5 = 0;

ll dfs(ll n){
    if(n == 0 || n == 1) return 1;
    ll tmp = n;
    while(n > 0 && n%2 == 0){
        cnt2++;
        n /= 2;
    }
    while(n > 0 && n%5 == 0){
        cnt5++;
        n /= 5;
    }
    tmp*dfs(tmp-2);
}

int main() {
    ll n;
    cin >> n;
    dfs(n);
    cout << min(cnt2,cnt5) << endl;
    return 0;
}