Problem: For given two segments s1 and s2, print the coordinate of the cross point of them.

s1 is formed by end points p0 and p1, and s2 is formed by end points p2 and p3.

Constraints

* 1 ≤ q ≤ 1000
* -10000 ≤ xpi, ypi ≤ 10000
* p0 ≠ p1 and p2 ≠ p3.
* The given segments have a cross point and are not in parallel.

Input

The entire input looks like:


q (the number of queries)
1st query
2nd query
...
qth query


Each query consists of integer coordinates of end points of s1 and s2 in the following format:


xp0 yp0 xp1 yp1 xp2 yp2 xp3 yp3


Output

For each query, print the coordinate of the cross point. The output values should be in a decimal fraction with an error less than 0.00000001.

Example

Input

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


Output

1.0000000000 0.0000000000
0.5000000000 0.5000000000
0.5000000000 0.5000000000

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;

typedef long long lint;
typedef long double louble;

template<typename T1,typename T2> inline T1 max(T1 a,T2 b){return a<b?b:a;}
template<typename T1,typename T2> inline T1 min(T1 a,T2 b){return a<b?a:b;}

namespace ae86
{
	const int bufl = 1<<15;

	char buf[bufl],*s=buf,*t=buf;

	inline int fetch()
	{
		if(s==t){t=(s=buf)+fread(buf,1,bufl,stdin);if(s==t)return EOF;}
		return *s++;
	}

	inline int ty()
	{
		int a=0,b=1,c=fetch();
		while(!isdigit(c))b^=c=='-',c=fetch();
		while(isdigit(c))a=a*10+c-48,c=fetch();
		return b?a:-a;
	}
}
using ae86::ty;

const double eps = 1e-11;

inline int dcmp(double x){if(-eps<x && x<eps)return 0;return x>0?1:-1;}

struct points
{
	double x,y;
	points(double _x=0,double _y=0){x=_x,y=_y;}

	inline friend points operator + (points a,points b){return points(a.x+b.x,a.y+b.y);}
	inline friend points operator - (points a,points b){return points(a.x-b.x,a.y-b.y);}
	inline friend points operator * (points a,double b){return points(a.x*b,a.y*b);}
	inline friend points operator / (points a,double b){return points(a.x/b,a.y/b);}

	inline friend int operator == (points a,points b){return dcmp(a.x-b.x)==0 && dcmp(a.y-b.y)==0;}

	void takein(){x=ty(),y=ty();}
	void print(char ends='\n'){printf("%.12lf %.12lf",x,y),putchar(ends);}
};

inline double dox(points a,points b){return a.x*b.x+a.y*b.y;}
inline double cox(points a,points b){return a.x*b.y-a.y*b.x;}
inline double length(points a){return sqrt(dox(a,a));}
inline double distan(points a,points b){return length(a-b);}

inline points pervec(points a){if(dcmp(length(a))<=0)return points(0,0);return a/length(a);}

inline int isponl(points p,points a,points b)
{
	return dcmp(cox(p-a,p-b))==0 && distan(a,p)<=distan(a,b) && distan(b,p)<=distan(a,b);
}

inline int iscrashed(points al,points ar,points bl,points br)
{
	double amix=min(al.x,ar.x),amxx=max(al.x,ar.x),amiy=min(al.y,ar.y),amxy=max(al.y,ar.y);
	double bmix=min(bl.x,br.x),bmxx=max(bl.x,br.x),bmiy=min(bl.y,br.y),bmxy=max(bl.y,br.y);
	
	if(amxx<bmix || bmxx<amix || amxy<bmiy || bmxy<amiy)return 0;
	if(isponl(al,bl,br) || isponl(ar,bl,br) || isponl(bl,al,ar) || isponl(br,al,ar))return 1;

	if(dcmp(cox(al-bl,br-bl))!=dcmp(cox(ar-bl,br-bl))
		&& dcmp(cox(bl-al,ar-al))!=dcmp(cox(br-al,ar-al)))return 1;
	return 0;
}

inline points crash(points al,points ar,points bl,points br)
{
	points a=ar-al,b=br-bl;
	double bas=cox(al-bl,b)/cox(b,a);
	return al+a*bas;
}

void finder()
{
	points al,ar,bl,br;
	al.takein(),ar.takein(),bl.takein(),br.takein();

	crash(al,ar,bl,br).print();
}

int main()
{
	int q=ty();
	while(q--)finder();
	
	return 0;
}