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: Accepted)

#include <stdio.h>
#include <math.h>
int main(void){
    // Your code here!
    long S,r;
    long sq,sa,x,y,yy;
    long Xa,Ya,Xb,Yb,Xc,Yc;
    
    scanf("%ld\n",&S);
    
    sq = sqrt(S);
    
    //printf("%ld %ld\n",sq,sa);
    int i,j;
    int flag=0;
    for(i=0;S>=i;i++){
        y = i;
        sa = (sq+i)*(sq+i) - S;
        if(sa==0) {x=0;yy=0;break;}
        for(j=1;sa>=j;j++){
            yy = j;
            if(sa%j==0){
                if(j==0) x = 0;
                else     x = sa/j;
                if(x<=1000000000){
                    flag = 1;
                    break;
                }
            }
        }
        if (flag==1) break;
    }
    //y = -1;
    //x = sa;
    
    Xa = 0;
    Ya = 0;
    Xb = x;
    Yb = sq+y;
    Xc = sq+y;
    Yc = yy;
    
    printf("%ld %ld %ld %ld %ld %ld\n",Xa,Ya,Xb,Yb,Xc,Yc);
    
    //確認
    //r = -(Xb-Xa)*(Yc-Ya)+(Xc-Xa)*(Yb-Ya);
    //printf("%ld\n",r);
    return 0;
}