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 <bits/stdc++.h>
#define endl '\n'
#define fast ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define zy -2147382637
#define bql 2147483647
#define ll long long
#define ull unsigned long long
#define ld long double
#define il inline
#define sz(x) x.size()
#define maxn 100010
#define rp(i, l, r) for (int i = l; i <= r; i++)
#define rb(i, r, l) for (int i = r; i >= l; i--)
#define fi first
#define se second
#define pb push_back
#define mp make_pair

using namespace std;
ll max(ll a,ll b){if(a>b)return a;else return b;}
ll min(ll a,ll b){if(a<b)return a;else return b;}
ll lowbit(ll x){return x&(-x);}
ll prime(ll x){
   if(x<=1)return false;
   for(int i=2;i<=int(sqrt(x));i++){
	   if(x%i==0)return false;
   }return true;
}
bool cmp(ll a,ll b){return a>b;}
ll gcd(ll a,ll b){ll r;while(b>0){r=a%b;a=b;b=r;}return a;}
ll powmod(ll a,ll b,ll mod) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
inline ll read() {
	char c = getchar(); ll w = 1; ll x = 0;
	while (!isdigit(c))
		(c == '-') && (w = -w), c = getchar();
	while (isdigit(c))
		x = (x << 1) + (x << 3) + (c ^ '0'), c = getchar();
	x *= w;
}
void fl(string name){
	freopen((name+".in").c_str(),"r",stdin);
	freopen((name+".out").c_str(),"w",stdout);
}
const int dx[8] = {-1, 1, 0, 0, -1, 1, -1, 1},
		  dy[8] = {0, 0, -1, 1, -1, -1, 1, 1};
int main(){
	fast;
	ll n;
	cin>>n;
	cout<<n/10<<endl;
	return 0;
}/*『你强归你强，我永不示弱』*/