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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;

public class CSharp
{
    public static void Main()
    {
        Kyopuro chokudai = new Kyopuro();
        chokudai.AtCoder();
    }
}

public class Kyopuro
{


    public void AtCoder()
    {
        string[] input;
        input = Console.ReadLine().Split(' ');
        long n = long.Parse(input[0]);
        
        int p = 998244353;
        int x = 0;
        while (x * x <= n) x++;
        x--;
        for(int i=x; i>0; i--)
        {
            if(n % x == 0)
            {
                Console.WriteLine
                    (0 + " " + 0 + " " + i + " " + 0 + " " + 0 + " " + (n / i));
                break;
            }
        }

        
    }
    public static void DebugWL(object x) { System.Diagnostics.Debug.WriteLine(x); }
}

// Mod(int n)
// int[] fac: factors (1-indexed)
// int n: the length of fac

// int .Ncr(n,r): nCr
// int .Fac(n): n!%p
// int .Pow(n,m): n^m%p

public class Mod
{
    const int p = 998244353;
    int[] fac;

    public Mod(int n)
    {
        fac = new int[n + 1];
        fac[0] = 1;
        long tmp = 0;
        for (int i = 1; i <= n; i++)
        {
            tmp = fac[i - 1];
            tmp *= i;
            tmp %= p;
            fac[i] = (int)tmp;
        }


    }

    public int Ncr(int n, int r)
    {
        if (n < r) return 0;
        if (r == 0 | n == r) return 1;
        long a = Fac(n);
        a = a * Pow(Fac(n - r), p - 2) % p;
        a = a * Pow(Fac(r), p - 2) % p;
        return (int)a;
    }

    public int Fac(int n)
    {
        if (fac[n] >= 0) return fac[n];
        long k = 1;
        for (long i = 1; i <= n; i++)
        {
            k = k * i % p;
        }
        fac[n] = (int)k;
        return fac[n];
    }

    public int Pow(int n, int m)
    {
        if (m == 0) return 1;
        if (m == 1) return n;
        long a = 1;
        if (m % 2 == 0)
        {
            a = Pow(n, m / 2);
            a = a * a % p;
        }
        else
        {
            a = Pow(n, m / 2);
            a = a * a % p;
            a = a * n % p;
        }
        return (int)a;
    }
}