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: Time Limit Exceeded)

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>

#define ll long long
#define pb push_back
#define sz(x) (int)(x).size()
#define mp make_pair
#define f first
#define s second
#define all(x) x.begin(), x.end()
#define D(x) cerr << #x << " is " << (x) << "\n";

using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>; ///find_by_order(),order_of_key()
template<class T1, class T2> ostream& operator<<(ostream& os, const pair<T1,T2>& a) { os << '{' << a.f << ", " << a.s << '}'; return os; }
template<class T> ostream& operator<<(ostream& os, const vector<T>& a){os << '{';for(int i=0;i<sz(a);i++){if(i>0&&i<sz(a))os << ", ";os << a[i];}os<<'}';return os;}
template<class T> ostream& operator<<(ostream& os, const set<T>& a) {os << '{';int i=0;for(auto p:a){if(i>0&&i<sz(a))os << ", ";os << p;i++;}os << '}';return os;}
template<class T> ostream& operator<<(ostream& os, const multiset<T>& a) {os << '{';int i=0;for(auto p:a){if(i>0&&i<sz(a))os << ", ";os << p;i++;}os << '}';return os;}
template<class T1,class T2> ostream& operator<<(ostream& os, const map<T1,T2>& a) {os << '{';int i=0;for(auto p:a){if(i>0&&i<sz(a))os << ", ";os << p;i++;}os << '}';return os;}

ll check=1e12;
int a1,a2;
bool checker(ll a)
{
    if(a==0)
    {
        return false;
    }
    if(a==1)
    {
        a1=a2=1;
        return false;
    }
    if(a>check)
        return true;
    for(ll i=2;i*i<=a;i++)
        if(a%i==0)
        {
            a1=i;
            a2=a/i;
            return false;
        }
    return true;
}
int main()
{
    ll s;
    scanf("%lld",&s);
    int sq=sqrt(s),lim=1e9;
    for(int i=sq;i<=lim;i++)
    {
        int m=s/i;
        if(s%i)
            m++;
        for(int j=m;j<=lim&&j-m<1000;j++)
        {
            ll diff=(ll)i*j-s;
            if(diff<0)
                continue;
            if(!checker(diff))
            {
                printf("0 0 %i %i %i %i\n",i,a1,a2,j);
                return 0;
            }
        }
    }
    while(true);
}