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>

#define sf scanf
#define pf printf
#define pb push_back
#define mp make_pair
#define PI ( acos(-1.0) )
#define mod 998244353LL
#define IN freopen("C.in","r",stdin)
#define OUT freopen("output.txt","w",stdout)
#define FOR(i,a,b) for(i=a ; i<=b ; i++)
#define DBG pf("Hi\n")
#define i64 long long int
#define eps (1e-8)
#define xx first
#define yy second
#define ln 42
#define off 2
#define FastIO  ios_base::sync_with_stdio(false); cin.tie(NULL)


using namespace __gnu_pbds;
using namespace std ;

typedef tree< i64, null_type, less<i64>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
typedef pair<i64, i64> pii;

#define maxn 100005

int main()
{
    i64 s ;

    scanf("%lld",&s) ;

    i64 sq = sqrt(s) ;
    sq = max( 0LL , sq-1 ) ;

    while( sq*sq < s ) sq++ ;

    if( sq*sq==s )
    {
        printf("0 0\n0 %lld\n%lld 0\n",sq,sq) ;
        return 0 ;
    }

    i64 mx = 1000000000LL ;

    if( s >= (sq-1)*(sq) )
    {
        printf("%lld 1\n",sq) ;
        printf("%lld %lld\n", sq*sq-s , sq ) ;
        printf("0 0\n") ;
    }
    else{
        printf("%lld 1\n",sq-1);
        printf("%lld %lld\n",(sq-1)*sq - s , sq  ) ;
        printf("0 0\n") ;
    }

    return 0 ;
}