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: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define REP(i,n) for(ll i=0, i##_len=(n); i<i##_len; ++i)
#define REPR(i,n) for(ll i = n;i >= 0;--i)
#define FOR(i,m,n) for(ll i = m, i##_len=(n);i <i##_len; ++i)
#define all(x) (x).begin(),(x).end()
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
const int INF = 1e9;
const ll LLINF = 1e16;
 
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;
}

int main(void)
{
    ll s;
    cin >> s;
    
    if(s == 1e18){
        cout << 0 << " " << 0 << " " << (ll)1e9 << " " << 1 << " " 
    <<  0 << " " << (ll)1e9 <<endl;
    }
    else{
        cout << 0 << " " << 0 << " " << (ll)1e9 << " " << 1 << " " 
    <<  (ll)1e9 - s % ((ll)1e9) << " " <<s / ((ll)(1e9)) + 1 <<endl;
    }

    return 0;
}