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)

//From Naruto to Hokage
//#include <RJ>
#include<bits/stdc++.h>
using namespace std;

#define ll          long long
#define pb          push_back
#define endl        '\n'
#define mii         map<ll,ll>
#define pii         pair<ll,ll>
#define vi          vector<ll>
#define all(a)      (a).begin(),(a).end()
#define F           first
#define S           second
#define sz(x)       (ll)x.size()
#define hell        1000000007
#define INF         (1ll<<60)
#define rep(i,a,b)  for(ll i=a;i<=b;i++)
#define rrep(i,a,b)  for(ll i=a;i>=b;i--)
#define ios ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define time        cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a % b); }
ll lcm(ll a, ll b) { return a * (b / gcd(a, b)); }
const int N=100005;
void solve(){
    ll s;
    cin>>s;
    ll x=1000000000;
    ll a=s/x;
    a++;
    ll b=a*x-s;
    cout<<"0 0 1000000000 1 "<<b<<" "<<a<<endl;
}
signed main()
{
    ios
    int TESTS=1;
    //cin>>TESTS;
    while(TESTS--){
        solve();
    }
    time
    return 0;
}