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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

							//	package Dynamic;
import java.io.*;
import java.util.*;
class Main{
    static public void main(String[] args) throws IOException{
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
//        int t=Integer.parseInt(br.readLine());
//        while(t-->0)
//        {
//        	int n=Integer.parseInt(br.readLine());
//        	int ar[]=new int[n];
//        	StringTokenizer st=new StringTokenizer(br.readLine());
        	//int a=Integer.parseInt(st.nextToken());
        	//int b=Integer.parseInt(st.nextToken());
        	//int c=Integer.parseInt(st.nextToken());
//        	for(int x=0;x<n;x++)
//        	{
//        		ar[x]=Integer.parseInt(st.nextToken());
//        	}
        long a=Long.parseLong(br.readLine());
        long v=(long)Math.sqrt(a);
        
        while(true)
        {
        	double d=(1.0*a)/v;
        	if(d==(double)((long)d))
        		break;
        	else
        		v--;
        }
        	System.out.println("0 "+v+" 0 0 "+(a/v)+" 0");
        //}
    }
}