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 <iostream>
#include <string>
#include <vector>
#include <math.h>
#include <algorithm>
#include <cmath>
#include <ios>
#include <ctype.h>
#include <stack>
#include <istream>
#include <stdio.h>
#include <map>
#include <set>
#include <queue>
#include <unordered_map>
#include <functional>
#include <bitset>
#include <list>
#include <deque>
#include <time.h>
#include <random>
#include <iomanip>
#include <fstream>
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define REP(i,a,b) for (int i = a; i < (b); ++i)
#define all(x) (x).begin(),(x).end()
#define pint pair<int,int>
const int INF = 1000000007;
const long double EPS = 1e-15;
const long double PI = acos(-1);
typedef long long ll;
using namespace std;
using Graph = vector<vector<int>>;
using Graph_weighted = vector<vector<pair<int,int>>>;
ll gcd(ll a, ll b) { return b ? gcd(b, a%b) : a; }
ll lcm(ll a, ll b) { return (a * b) / gcd(a,b); }
ll ceil(const ll a, const ll b) { return (a + b - 1) / b; }
int main(){
  ll s;
  cin >> s;
  ll a = 0,b = 0;
  for(ll i = sqrt(s) + 1; i >= 0; i--){
    if(s % i == 0){
      a = i,b = (s / i);
      break;
    }
  }
  ll c = min(a,b),d = max(a,b);
  cout << 0 << " " << 0 << " " << c << " " << 0 << " " << c << " " << d << endl;
  return 0;
}