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 <bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i, a, b) for(int i = a; i < b; i++)
#define rrep(i, a, b) for(int i = b - 1; i >= a; i--)
#define ALL(a) a.begin(), a.end()
#define pii pair<int,int>
#pragma GCC optimize("Ofast")
#define pcnt __builtin_popcount
#define buli(x) __builtin_popcountll(x)
#define pb push_back
#define mp make_pair
#define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() );
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(a>b){a = b; return 1;} return 0; };
inline void IN(void){return;}
template <typename First, typename... Rest> void IN(First& first, Rest&... rest){cin >> first;IN(rest...);return;}
inline void OUT(void){cout << "\n";return;}
template <typename First, typename... Rest> void OUT(First first, Rest... rest){cout << first << " ";OUT(rest...);return;}
const double EPS = 1e-9;
const int mod = 1e9 + 7;
const int INF = 1e9;
const long long LLINF = 1e18;
long long lcm(ll a, ll b){return a * b / __gcd(a,b);}
struct IoSetup { IoSetup() {
    cin.tie(nullptr);ios::sync_with_stdio(false);
    cout << fixed << setprecision(10);
    cerr << fixed << setprecision(10);
} } iosetup;
template< int mod > struct ModInt {
    int x; ModInt() : x(0) {}
    ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}
    ModInt &operator+=(const ModInt &p) {if((x += p.x) >= mod) x -= mod;return *this;}
    ModInt &operator-=(const ModInt &p) {if((x += mod - p.x) >= mod) x -= mod;return *this;}
    ModInt &operator*=(const ModInt &p) {x = (int) (1LL * x * p.x % mod);return *this;}
    ModInt &operator/=(const ModInt &p) {*this *= p.inverse();return *this;}
    ModInt operator-() const { return ModInt(-x); }
    ModInt operator+(const ModInt &p) const { return ModInt(*this) += p; }
    ModInt operator-(const ModInt &p) const { return ModInt(*this) -= p; }
    ModInt operator*(const ModInt &p) const { return ModInt(*this) *= p; }
    ModInt operator/(const ModInt &p) const { return ModInt(*this) /= p; }
    bool operator==(const ModInt &p) const { return x == p.x; }
    bool operator!=(const ModInt &p) const { return x != p.x; }
    ModInt inverse() const {int a = x, b = mod, u = 1, v = 0, t;
    while(b > 0) { t = a / b; swap(a -= t * b, b); swap(u -= t * v, v); }return ModInt(u);}
    ModInt pow(int64_t n) const {ModInt ret(1), mul(x); while(n > 0) {if(n & 1) ret *= mul;mul *= mul;n >>= 1;}return ret;}
    friend ostream &operator<<(ostream &os, const ModInt &p) { return os << p.x;}
    friend istream &operator>>(istream &is, ModInt &a) { int64_t t; is >> t; a = ModInt< mod >(t); return (is); }
    static int get_mod() { return mod; }
}; using modint = ModInt< mod >;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};

int main(){
    iosetup;
    ll S; cin >> S;
    ll x = S % INF;
    ll y = S / INF;
    if(y == INF) cout << 0 << " " << 0 << " " << INF << " " << 0 << " " << 0 << " " << INF << endl;
    else cout << 0 << " " << 0 << " " << INF << " " << 1 << " " << INF - x << " " << 1 + y << endl;
    return 0;
}