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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

#ifndef bhartiya
    #pragma GCC optimize("Ofast")
    #pragma GCC optimize("unroll-loops")
    // #pragma GCC target ("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#endif

#define int long long
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef gp_hash_table<int, int> umap;
typedef tree<int, null_type, less<int>, rb_tree_tag,
		tree_order_statistics_node_update> oset;
// not that imp
typedef pair<pii, int> piii;
typedef vector<int> vi;
typedef vector<pii> vii;

#define inf 200000000000000
#define mod 1000000007
#define eps 1e-7
#define PI 3.1415926535897932385
// #define PI acos(-1)

#define pb push_back
#define bitc  __builtin_popcountll
#define mp make_pair
#define ff first
#define ss second
#define all(ar) ar.begin(), ar.end()
#define sz(x) (int)(x).size()

#define fr(i,a,b) for (int i = (a), _b = (b); i <= _b; i++)
#define rep(i,n) for (int i = 0, _n = (n); i < _n; i++)
#define repr(i,n) for (int i = n - 1; i >= 0; i--)
#define frr(i,a,b) for (int i = (a), _b = (b); i >= _b; i--)
#define foreach(it,ar) for (auto it = ar.begin(); it != ar.end(); it++)
#define fil(ar,val) memset(ar, val, sizeof(ar))

#ifdef bhartiya
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
	cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
	const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif

mt19937 rng32(chrono::steady_clock::now().time_since_epoch().count());
inline bool equals(double a, double b) {return fabs(a - b) < 1e-9;}
int gcd(int a, int b){ return b==0 ? a : gcd(b, a%b); }
ll modpow(ll b, ll e) {
	ll ans=1; for(;e;b=b*b%mod,e/=2) if(e&1) ans=ans*b%mod; return ans;}

const int maxn = 1000005;

ld dist (int x, int y, int x2, int y2){
	return sqrt((x-x2)*(x-x2) + (y-y2)*(y-y2));
}



void solve(){
	int n; cin>>n;
	ld z = sqrt(n);
	int x = ceil(z);
	int y = ceil(z)*ceil(z);
	int diff = y - n;
	cout<<"0 0 "<<x<<" "<<1<<" "<<diff<<" "<<x;
}

signed main() 
{ 
	ios_base::sync_with_stdio(false);
    cin.tie(0), cout.tie(0), cout.precision(10); //cout<<fixed;
    cin.exceptions(cin.failbit);
    #ifdef bhartiya
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout);
    #endif
    solve();
    end_routine();
}