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;
#define pa pair<int, int> 
#define pal pair<long long, long long>
#define pali pair<long long, int>
#define pad pair<double, double>
#define pb push_back
#define mp make_pair 
#define COUT(v) for(int64_t i=0;i<(v).size();i++){ cout << v.at(i) << endl;}
#define REP(i, n) for(int64_t i=0;i<n;i++)
#define VIN(v) for(int64_t i=0;i<(v).size();i++){cin >> (v).at(i);}

typedef vector<bool> bvec;
typedef vector<int> ivec;
typedef vector<long long> lvec;
typedef vector<double> dvec;
typedef vector<pa> pavec;
typedef vector<pali> palivec;
typedef vector<pal>palvec;


typedef vector< vector<bool> > bmat;
typedef vector< vector<int> > imat;
typedef vector< vector<long long> > lmat;

typedef vector<string> svec;
typedef vector< vector<string> > smat;
const ll infll = (1LL << 60) - 1;
const int inf = (1 << 30) - 1;
const int MOD = 1000000007;


ll gcd(ll x, ll y){
    ll r = x%y;
    if(r == 0) return y;
    else return gcd(y, r);
} 

ll lcm(ll x, ll y){
    return x*y/gcd(x,y);
}

lvec mfactor(ll n){
    bvec ip(n, true);
    lvec mf(n, -1);
    ip[0]=false; ip[1]=false; mf[0]=0; mf[1]=1;
    REP(i, n){
        if(ip[i]){
            mf[i] = i;
            for(ll j=i*i; j<n; j+=i){
                ip[j] = false;
                if(mf[j] == -1) mf[j]=i;
            }
        }
    }
    return mf;
}

palivec get_prime(ll n, const lvec &mf){
    palivec plist;
    while(n != 1){
        int cnt=0;
        ll m=mf[n];
        while(mf[n] == m){
            cnt++;
            n/=m;
        }
        plist.pb(pali(m, cnt));
    }
    return plist;
}

void COMinit(int m, lvec &fac, lvec &finv){
    lvec inv(m);
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < m; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

ll COM(int n, int k, const lvec& fac, const lvec &finv){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}


ll modinv(ll a, ll m) {
    ll b = m, u = 1, v = 0;
    while (b) {
        ll t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}

ll powmod(ll x, ll n){
    ll res=1;
    while(n>0){
        if(n&1) res = res*x%MOD;
        x = x*x%MOD;
        n>>=1;
    }
    return res;
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll n;
    cin >> n;
    if(n%2==1) {
        cout << 0<< endl;
        return 0;
    }
    ll x=10;
    ll ans=0;
    while(n/x > 0 ){
        ans += n/x;
        x*=5;
    }
    cout << ans << endl;
}