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;
const ll INF = 1LL<<60;
#define all(x) (x).begin(),(x).end()
#define REP(i,n) for(int i=0; i<(n);++i)
#define FOR(i,k,n) for(int i=(k);i<(n);++i)

int cnt5(ll x){
    int res = 0;
    ll tmp = 1;
    while(tmp<x){
        tmp*=5;
        res++;
    }
    return res;
}

ll pow5(int k){
    ll res = 1;
    REP(i,k) res *= 5;
    return res;
}

int main(){
    ll N; cin >> N;
    ll ans = 0;
    FOR(i,1,cnt5(N)+1){
        ans += N/2/pow5(i);
    }
    cout << (N%2==0?ans:0) << endl;
    return 0;
}