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>
#define N 1000000000

int main(void){
    unsigned long long int s,t;
    unsigned long long int res[6] = {};
    int i,j;
    
    scanf("%llu", &s);
    t = s;
    i = 0;
    while(t){
        t >>= 1;
        i++;
    }
    t = s;
    t >>= (i-1)/2;
    
    for(i = t; i * t > s; i--){
        ;
    }
    
    res[2] = i+1;
    res[5] = t;
    if(res[2] * res[5] != s){
        t = res[2] * res[5] - s;
        //printf("t:%llu\n", t);
        i = t;
        j = 0;
        while(i){
            i >>= 1;
            j++;
        }
        i = t;
        i >>= (j-1)/2;
        if(i == 0){
            i = 1;
        }else{
           while(i > 1){
                if(t % i == 0){
                    break;
                }
                i--;
            } 
        }
        
        
        res[3] = t / i;
        res[4] = i;
    }
    
    for(i = 0; i < 6; i++){
        printf("%llu ", res[i]);
    }
    printf("\n");
    
    return 0;
}