Problem:
Score : 400 points

Problem Statement

Given is an integer S.
Find a combination of six integers X_1,Y_1,X_2,Y_2,X_3, and Y_3 that satisfies all of the following conditions:

0 \leq X_1,Y_1,X_2,Y_2,X_3,Y_3 \leq 10^9

The area of the triangle in a two-dimensional plane whose vertices are (X_1,Y_1),(X_2,Y_2), and (X_3,Y_3) is S/2.

We can prove that there always exist six integers that satisfy the conditions under the constraints of this problem.

Constraints

1 \leq S \leq 10^{18}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

S

Output

Print six integers X_1,Y_1,X_2,Y_2,X_3, and Y_3 that satisfy the conditions, in this order, with spaces in between.
If multiple solutions exist, any of them will be accepted.

Sample Input 1

3

Sample Output 1

1 0 2 2 0 1

The area of the triangle in a two-dimensional plane whose vertices are (1,0),(2,2), and (0,1) is 3/2.
Printing 3 0 3 1 0 1 or 1 0 0 1 2 2 will also be accepted.

Sample Input 2

100

Sample Output 2

0 0 10 0 0 10

Sample Input 3

311114770564041497

Sample Output 3

314159265 358979323 846264338 327950288 419716939 937510582

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
using ll = long long;

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

#define FOR(i,a,b) for(ll i=(a);i<(b);++i)
#define ALL(v) (v).begin(), (v).end()
#define p(s) cout<<(s)<<endl
#define p2(s, t) cout << (s) << " " << (t) << endl
#define br() p("")
#define pn(s) cout << (#s) << " " << (s) << endl
#define p_yes() p("YES")
#define p_no() p("NO")

const ll mod = 1e9 + 7;
const ll inf = 1e18;

// 素数判定
bool isPrime(ll n){
    if (n < 2) return false;
    else if (n == 2) return true;
    else if (n % 2 == 0) return false;

    double sqrtNum = sqrt(n);
    for (ll i=3; i<=sqrtNum; i+=2){
        if (n%i == 0){
            return false;
        }
    }
    return true;
}




void vprint(vector<ll> A){
    ll L = A.size();
    FOR(i, 0, L){
        if(i) cout << ' ';
        cout << A[i];
    }
    cout << endl;
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    // input
    ll S;
    cin >> S;

    ll x1 = 1e9;
    ll x2 = 1;

    ll y2 = S/1e9;
    ll y1 = x1 * y2 - S;

    cout << "0 0 " << x1 << ' ' << y1 << ' ' << x2 << ' ' << y2 << endl;
    
    return 0;
}