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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <math.h>

int main(){
  long int s;
  scanf( "%ld", &s );
  //printf( "s: %ld\n", s );
  
  int x1, y1, x2, y2, x3, y3;
  x1 = 0, y1 = 0, y2 = 0, x3 = 0;
  int i_x2 = 0;
  long int h = 0, w = 0;

  // if ( s > 1000000000000000000 / 2 ){
  //   for ( i_x2 = 1000000000; i_x2 >= 1; i_x2-- ){
  //     w = i_x2;
  //     h = s / i_x2;
  //     if ( h > 1000000000 ) continue;
  //     long int o = sqrt( ( long int )w * w + ( long int )h * h );
  //     if ( ( h + w <= o ) || ( h + w <= o ) || ( h + w <= o ) ) continue;
  //     if ( ( long int )w * h == s ) break;

  //   }
  // }else { 
  //   for ( i_x2 = 1; i_x2 <= 1000000000; i_x2++ ){
  //     w = i_x2;
  //     h = s / i_x2;

  //     if ( h > 1000000000 ) continue;

  //     long int o = sqrt( ( long int )w * w + ( long int )h * h );
  //     if ( ( long int )w * h == s ) break;

  //   }
  // }
  if ( s > 1000000000000000000 / 2 ){
    for ( i_x2 = sqrt( s ); ( long int )i_x2 * i_x2 <= s; i_x2-- ){
        if ( s % ( long int )i_x2 == 0 ){
          h = s / i_x2;
          if ( h > 1000000000 ) continue;
          break;
        }
     }
  }else{
    for ( i_x2 = 1; ( long int )i_x2 * i_x2 <= s; i_x2++ ){
      if ( s % ( long int )i_x2 == 0 ){
        h = s / i_x2;
        if ( h > 1000000000 ) continue;
        break;
      }
    }
  }
  x2 = i_x2;
  y3 = h;

  printf( "%d %d %d %d %d %d\n", x1, y1, x2, y2, x3, y3 );

  return 0;
}