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 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
#include <inttypes.h>

int main(void) {
	int64_t S;
	int64_t smaller = 0, ge = INT64_C(1000000000);
	int64_t gosa1, gosa2;
	if (scanf("%" SCNu64, &S) != 1) return 1;
	while (smaller + 1 < ge) {
		int64_t m = smaller + (ge - smaller) / 2;
		if (S <= m * m) ge = m; else smaller = m;
	}
	gosa1 = S - smaller * smaller;
	gosa2 = ge * ge - S;
	if (gosa2 <= gosa1) {
		printf("0 0 %" PRId64 " 1 %" PRId64 " %" PRId64 "\n", ge, gosa2, ge);
	} else {
		printf("0 1 %" PRId64 " 0 %" PRId64 " %" PRId64 "\n", smaller, gosa1, smaller + 1);
	}
}

/*

1点を原点に固定してよい。 (多分)

【応用】ベクトルの内積と三角形の面積 | なかけんの数学ノート
https://math.nakaken88.com/textbook/expert-inner-product-of-vector-and-triangle-area/

S/2 = abs(x1*y2 - y1*x2) / 2

x = 10**9 とすると、

  x
= 1000000000
  (x*x) - ((x-1)*(x-1))
= 1999999999

なので、誤差を規定内に抑えられるはず

x1 = y2 = 一番近くなるやつ
y1 = 1
x2 = 誤差

誤差が負になるとき、y1 = -1 にする → 全体を1上げる

*/