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;

#define REP(i, n) for (int i = (int)(0); i < (int)(n); ++i)
#define REPS(i, n) for (int i = (int)(1); i <= (int)(n); ++i)
#define RREP(i, n) for (int i = ((int)(n)-1); i >= 0; i--)
#define RREPS(i, n) for (int i = ((int)(n)); i > 0; i--)
#define IREP(i, m, n) for (int i = (int)(m); i < (int)(n); ++i)
#define IREPS(i, m, n) for (int i = (int)(m); i <= (int)(n); ++i)
#define FOR(e, c) for (auto &e : c)
#define SORT(v, n) sort(v, v + n);
#define VSORT(v) sort(v.begin(), v.end());
#define RVISORT(v) sort(v.begin(), v.end(), greater<int>());
#define ALL(v) v.begin(), v.end()
#define MP(n, m) make_pair(n, m);

using VI = vector<int>;
using VVI = vector<VI>;
using PII = pair<int, int>;
using ll = long long;
using ul = unsigned long;

typedef long long ll;

template<class T, class C> void chmax(T& a, C b){ a>b?:a=b; }
template<class T, class C> void chmin(T& a, C b){ a<b?:a=b; }

const int mod=1e9+7;

void printv(VI& v){
  REP(i,v.size()) cout << v[i] << " ";
  cout << endl;
}

int vtotal(VI& v){
  int total=0;
  REP(i,v.size()) total+=v[i];
  return total;
}

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

	ll s;
	cin>>s;
  int lim=1e9;
  int y=(s/lim)+1;
  int x=(ll)y*lim-s;

  cout << 0 << " " << 0 << " " << lim << " " << 1 << " " << x << " " << y << endl;

  return 0;
}