/* Tenka1 Programmer Contest E
   首先,考虑到曼哈顿距离=|xi-xj|+|yi-yj|,
   由这样的性质,可以得到x,y是相互独立的.
   问题被转化为一维的情况,
   现在就是对k*(k-1)/2个点,需要去寻找一个值k使得sigma(abs(k-xi))最小,
   由中位数的性质,显然,k就是这k*(k-1)/2个点的中位数.
   考虑二分答案,现在就是要求对于每个k统计出x坐标小于k的点的数量.
   将所有线按照斜率排序,当x很小时,用一条竖直线段去切这n条线,斜率小的位置最靠下,而斜率大的位置最靠上,
   当x=k时,如果在x<k时l1和l2发生了相交,则l1和l2的相对顺序就会发生变化,
   于是问题就转化为了求逆序对的数量,这是O(nlogn)复杂度的,
   于是就可以得到一个O(nlognlog(eps^(-1)))的解法.
*/
#include <bits/stdc++.h>
#define lim 200000005
using namespace std;
struct line
{double a,b,c,k;
int id;
}p[40005];
inline bool cmp(line p,line q)
{return p.k<q.k;}
int n,all,nd,t[40005];
int e[40005];
inline void add(int pos)
{while (pos<=n)
{e[pos]++;pos+=(pos&(-pos));}
}
inline int ask(int pos)
{int s=0;
while (pos)
{s+=e[pos];pos-=(pos&(-pos));}
return s;
}
inline int calc(double mid)
{int i;
for (i=1;i<=n;i++)
{p[i].k=(-p[i].a/p[i].b)*mid+p[i].c/p[i].b;}
sort(p+1,p+n+1,cmp);
for (i=1;i<=n;i++)
{t[i]=p[i].id;}
int ans=0;
memset(e,0,sizeof(e));
for (i=1;i<=n;i++)
{ans+=ask(n)-ask(t[i]-1);
add(t[i]);
}
return ans;
}
inline void solve()
{int i,tms=66;
for (i=1;i<=n;i++)
{p[i].id=n-i+1;}
double l=-lim,r=lim;
while (tms--)
{double mid=(l+r)/2.0;
if (calc(mid)<nd) {l=mid;}
else {r=mid;}
}
printf ("%.12lf",l);
return;
}
int main (){
	int i;
	scanf ("%d",&n);
	all=n*(n-1)/2;
	if (all&1) {nd=(all+1)/2;}
	else {nd=all/2;}
	for (i=1;i<=n;i++)
	{scanf ("%lf%lf%lf",&p[i].a,&p[i].b,&p[i].c);
	p[i].k=-p[i].a/p[i].b;
	//a*x+b*y=c---->y=-a/b*x+c/b
	}
	sort(p+1,p+n+1,cmp);
	solve();
	printf (" ");
	for (i=1;i<=n;i++)
	{double tp=p[i].a;
	p[i].a=p[i].b;
	p[i].b=tp;
	p[i].k=-p[i].a/p[i].b;
	}
	sort(p+1,p+n+1,cmp);
	solve();
	printf ("\n");
	return 0;
}
	
	
	

### Prompt

Develop a solution in Cpp to the problem described below:

There are N lines in the xy-plane. The i-th line is represented by A_ix+B_iy=C_i. Any two lines among the N+2 lines, the above N lines plus the x-axis and y-axis, cross each other at exactly one point.

For each pair 1 \leq i < j \leq N, there is a car at the cross point of the i-th and j-th lines. Even where three or more lines intersect at a point, a car is individually placed for each pair of lines. That is, there will be k(k-1)/2 cars placed at the intersection of k lines.

Those cars are already very old, and can only be moved parallel to the x-axis or y-axis.

Takahashi will hold an exhibition of antique cars at a place on the xy-plane. In order to avoid damaging the half-broken cars too much, he will select the place of the exhibition so that the total distance covered will be minimized when all the cars are moved to the place. If such a place is not uniquely determined, among the places that satisfy the condition above, the place with the minimum x-coordinate will be selected. If the place is still not uniquely determined, among the places that satisfy the two conditions above, the place with the minimum y-coordinate will be selected.

Find the place of the exhibition that will be selected.

Constraints

* 2 \leq N \leq 4 × 10^4
* 1 \leq |A_i|,|B_i| \leq 10^4(1 \leq i \leq N)
* 0 \leq |C_i| \leq 10^4(1 \leq i \leq N)
* No two given lines are parallel.
* All input values are integers.

Inputs

Input is given from Standard Input in the following format:


N
A_1 B_1 C_1
:
A_N B_N C_N


Outputs

Print the x-coordinate and y-coordinate of the place of the exhibition that will be selected, in this order, with a space in between. The output will be judged as correct when the absolute or relative error is at most 10^{-9}.

Examples

Input

3
1 1 1
2 -1 2
-1 2 2


Output

1.000000000000000 1.000000000000000


Input

4
1 1 2
1 -1 0
3 -1 -2
1 -3 4


Output

-1.000000000000000 -1.000000000000000


Input

7
1 7 8
-2 4 9
3 -8 -5
9 2 -14
6 7 5
-8 -9 3
3 8 10


Output

-1.722222222222222 1.325000000000000

### Response

```cpp
/* Tenka1 Programmer Contest E
   首先,考虑到曼哈顿距离=|xi-xj|+|yi-yj|,
   由这样的性质,可以得到x,y是相互独立的.
   问题被转化为一维的情况,
   现在就是对k*(k-1)/2个点,需要去寻找一个值k使得sigma(abs(k-xi))最小,
   由中位数的性质,显然,k就是这k*(k-1)/2个点的中位数.
   考虑二分答案,现在就是要求对于每个k统计出x坐标小于k的点的数量.
   将所有线按照斜率排序,当x很小时,用一条竖直线段去切这n条线,斜率小的位置最靠下,而斜率大的位置最靠上,
   当x=k时,如果在x<k时l1和l2发生了相交,则l1和l2的相对顺序就会发生变化,
   于是问题就转化为了求逆序对的数量,这是O(nlogn)复杂度的,
   于是就可以得到一个O(nlognlog(eps^(-1)))的解法.
*/
#include <bits/stdc++.h>
#define lim 200000005
using namespace std;
struct line
{double a,b,c,k;
int id;
}p[40005];
inline bool cmp(line p,line q)
{return p.k<q.k;}
int n,all,nd,t[40005];
int e[40005];
inline void add(int pos)
{while (pos<=n)
{e[pos]++;pos+=(pos&(-pos));}
}
inline int ask(int pos)
{int s=0;
while (pos)
{s+=e[pos];pos-=(pos&(-pos));}
return s;
}
inline int calc(double mid)
{int i;
for (i=1;i<=n;i++)
{p[i].k=(-p[i].a/p[i].b)*mid+p[i].c/p[i].b;}
sort(p+1,p+n+1,cmp);
for (i=1;i<=n;i++)
{t[i]=p[i].id;}
int ans=0;
memset(e,0,sizeof(e));
for (i=1;i<=n;i++)
{ans+=ask(n)-ask(t[i]-1);
add(t[i]);
}
return ans;
}
inline void solve()
{int i,tms=66;
for (i=1;i<=n;i++)
{p[i].id=n-i+1;}
double l=-lim,r=lim;
while (tms--)
{double mid=(l+r)/2.0;
if (calc(mid)<nd) {l=mid;}
else {r=mid;}
}
printf ("%.12lf",l);
return;
}
int main (){
	int i;
	scanf ("%d",&n);
	all=n*(n-1)/2;
	if (all&1) {nd=(all+1)/2;}
	else {nd=all/2;}
	for (i=1;i<=n;i++)
	{scanf ("%lf%lf%lf",&p[i].a,&p[i].b,&p[i].c);
	p[i].k=-p[i].a/p[i].b;
	//a*x+b*y=c---->y=-a/b*x+c/b
	}
	sort(p+1,p+n+1,cmp);
	solve();
	printf (" ");
	for (i=1;i<=n;i++)
	{double tp=p[i].a;
	p[i].a=p[i].b;
	p[i].b=tp;
	p[i].k=-p[i].a/p[i].b;
	}
	sort(p+1,p+n+1,cmp);
	solve();
	printf ("\n");
	return 0;
}
	
	
	

```