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: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
#define INTMAX 2147483647

int main()
{
    long s,x1,y1,x2,y2,x3,y3;
    int i, j, k;
    scanf("%ld", &s);
    if(s<=1000000000)
    {
        x1 = 0;
        y1 = 0;
        x2 = 1;
        y2 = 0;
        x3 = 0;
        y3 = s;
        printf("%d %d %d %d %d %d", x1, y1, x2, y2, x3, y3);
        return 0;
    }
    int st = (int)sqrt(s);
    //printf("%d\n", st);
    // for (i = st; i <= 1000000000;i++)
    // {
    //     if (s % i == 0)
    //     {
    //         x1 = 0;
    //         y1 = 0;
    //         x2 = i;
    //         y2 = 0;
    //         x3 = 0;
    //         y3 = s/i;
    //         printf("%d %d %d %d %d %d", x1, y1, x2, y2, x3, y3);
    //         return 0;
    //     }
    // }

    int r = s % st;
    x1 = 0;
    y1 = 0;
    x2 = st;
    y2 = st-r;
    x3 = 1;
    y3 = (s+st-r)/st;
    printf("%d %d %d %d %d %d", x1, y1, x2, y2, x3, y3);
            return 0;
    //printf("aaa");
}