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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#if defined(__GNUC__)
#include <bits/stdc++.h>
// __GNUC__
#else
#include "bits/stdc++.h"
#endif
using namespace std;

#define int long long

#if defined(__GNUC__)
#define __int32 __int32_t
#endif // __GNUC__

#define itn __int32
#define REP(i, n) for(itn (i) = 0; (i) < (n); ++(i))
#define REP2(i, x, n) for(itn (i) = x; (i) < (n); ++(i))
#define ALL(x) (x).begin(), (x).end()
#define RALL(x) (x).rbegin(), (x).rend()
#define SORT(x) std::sort(ALL(x))
#define RSORT(x, T) std::sort(ALL(x), greater<T>())
#define REVERCE(x) std::reverce(ALL(x))
#define ACM(x, y) std::accumulate(ALL(x), y)

#define INT_POW(x, y) [](int a, int b) -> int {int c = a; REP(i, b - 1){c *= a;} return c;}(x, y)
#define INT_POWM(x, y, m) [](int a, int b, int m) -> int {int c = a; REP(i, b - 1){c *= a; c %= m;} return c;}(x, y, m)

itn main() {
	std::ios::sync_with_stdio(false);

	int s;

	cin >> s;

	int v, h;

	h = v = sqrtl((long double)s);

	while (v * h <= s) {
		cout << v << ", " << h << '\n';
		++h;
	}

	int dis = v * h - s;

	cout << 0 << ' ' << 0 << ' ' << 1 << ' ' << v << ' ' << h << ' ' << dis;

	return 0;
}