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: Runtime Error)

# include <bits/stdc++.h>
# define rep(i, n) for(int i=0, i##_len=(n); i<i##_len; ++i)
# define reps(i, n) for(int i=1, i##_len=(n); i<=i##_len; ++i)
# define rrep(i, n) for(int i=((int)(n)-1); i>=0; --i)
# define rreps(i, n) for(int i=((int)(n)); i>0; --i)
# define ALL(x) (x).begin(), (x).end()
# define SZ(x) ((int)(x).size())
# define pb push_back
# define optimize_cin() cin.tie(0); ios::sync_with_stdio(false)
# define MSG(x) std::cout << #x << " : " << x << "\n";
using namespace std;
using lint = long long;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

int main()
{
    string S;
    cin >> S;
    cout << S << endl;


    if (S == "1000000000000000000")
    {
        cout << "0 0 1000000000 0 0 1000000000" << endl;
        return 0;
    }

    string s1 = S.substr(0, 9);
    string s2 = S.substr(9, 9);
    lint high = stoi(s1);
    lint low  = stoi(s2);

    cout << (lint)pow(10, 9) * high + low << endl;

    cout << "0 0 " << (lint)pow(10, 9) << " " << low << " 1 " << high << endl;

    return 0;
}