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: Accepted)

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cassert>
#include <iostream>
#include <sstream>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <utility>
#include <numeric>
#include <algorithm>
#include <bitset>
#include <complex>
#include <array>
#include <list>
#include <stack>
#include <valarray>

using namespace std;

typedef unsigned uint;
typedef long long Int;
typedef unsigned long long UInt;

const int INF = 1001001001;
const Int INFLL = 1001001001001001001LL;

template<typename T> void pv(T a, T b) { for (T i = a; i != b; ++i) cout << *i << " "; cout << endl; }
template<typename T> void chmin(T& a, T b) { if (a > b) a = b; }
template<typename T> void chmax(T& a, T b) { if (a < b) a = b; }
int in() { int x; scanf("%d", &x); return x; }
double fin() { double x; scanf("%lf", &x); return x; }
Int lin() { Int x; scanf("%lld", &x); return x; }

int main() {
  const Int S = lin();

  const Int X1 = 1000000000;
  const Int Y2 = (S + X1 - 1) / X1;
  const Int R = X1 * Y2 - S;

  if (R == 0) {
    printf("0 0 %lld 0 0 %lld\n", X1, Y2);
  } else {
    printf("0 0 %lld %lld 1 %lld\n", X1, R, Y2);
  }
  return 0;
}