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;
#define inf 1000000000
#define INF 1000000000000000
#define ll long long
#define ull unsigned long long
#define M (int)(1e9+7)
#define P pair<int,int>
#define PLL pair<ll,ll>
#define FOR(i,m,n) for(int i=(int)m;i<(int)n;i++)
#define RFOR(i,m,n) for(int i=(int)m;i>=(int)n;i--)
#define rep(i,n) FOR(i,0,n)
#define rrep(i,n) RFOR(i,n,0)
#define all(a) a.begin(),a.end()
#define IN(a,n) rep(i,n){ cin>>a[i]; }
const int vx[4] = {0,1,0,-1};
const int vy[4] = {1,0,-1,0};
#define PI 3.14159265
#define F first
#define S second
#define PB push_back
#define EB emplace_back
#define int ll
#define vi vector<int>
#define m (int)1e9



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

  int s;
  cin>>s;

  if(s==(int)1e18){
    cout<<0<<' '<<0<<' ';
    cout<<0<<' '<<m<<' ';
    cout<<m<<' '<<m<<endl;
    return 0;
  }


  cout<<0<<' '<<0<<' ';
  cout<<m<<' '<<m-s%m<<' ';
  cout<<1<<' '<<s/m+1<<endl;
}