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)

/*Lucky_Glass*/
#include<vector>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int N=700;
int n,mod,Nf,Ng;
vector< pair<int,int> > Epr;
int Ef[N+3][N+3];

int main(){
	scanf("%d%d",&n,&mod);
	Nf=2*n-1;Ng=2*n-1;
	for(int i=0;i<2*n;i++){
		while(i*i+Nf*Nf>4*n*n && Nf>=0) Nf--;
		while(i*i+Ng*Ng>=n*n && Ng>=0) Ng--;
		if(Ng==-1) Epr.push_back(make_pair(Nf+1,0));
		else  Epr.push_back(make_pair(Ng+1,Nf+1));
//		printf("%d %d\n",Epr.back().first,Epr.back().second);
	}
	sort(Epr.begin(),Epr.end());
	int Vans=0;long long typ=1;
	for(int k=0;k<=n;k++){
		memset(Ef,0,sizeof Ef);
		Ef[0][0]=1;
		for(int i=0,Ctf=0,Ctg=0;i<(int)Epr.size();Ctg+=Epr[i].second==0,Ctf+=Epr[i].second!=0,i++){
			for(int j=0;j<=k;j++){
				int tmp=Ef[i][j];
				if(Epr[i].second){
					if(j!=k) Ef[i+1][j+1]=(Ef[i+1][j+1]+1ll*tmp*(Epr[i].first-j-Ctg)%mod)%mod;
					Ef[i+1][j]=(Ef[i+1][j]+1ll*tmp*(Epr[i].second-k-n-(Ctf-j))%mod)%mod;
				}
				else
					Ef[i+1][j]=(Ef[i+1][j]+1ll*tmp*(Epr[i].first-j-Ctg)%mod)%mod;
			}
//			printf("%d %d\n",Ctf,Ctg);
		}
//		printf("%d\n",Ef[Epr.size()][k]);
		Vans=(Vans+typ*Ef[Epr.size()][k]+mod)%mod;
		typ*=-1;
	}
	printf("%d\n",Vans);
	return 0;
}