#include<iostream>
#include<complex>
#include<array>
#include<vector>
#include<cstdio>
#include<cassert>
using namespace std;

const double EPS = 1e-10;
template<class T>
bool eq(const T& a, const T& b) {return abs(a - b) < EPS;}

typedef  complex<double> point;

double cross(const point& a, const point& b) {
	return imag(conj(a)*b);
}
double dot(const point& a, const point& b) {
	return real(conj(a)*b);
}
int ccw(point a, point b, point c) {
	b -= a; c -= a;
	if (cross(b, c) > 0)   return +1;  // counter clockwise
	if (cross(b, c) < 0)   return -1;  // clockwise
	if (dot(b, c) < 0)     return +2;  // c--a--b on line
	if (norm(b) < norm(c)) return -2;  // a--b--c on line
	return 0;  // a--c--b on line or a----bc
}

typedef array<point, 2> line;

point projection(const line& l, const point& p) {
	double t = dot(p-l[0], l[0]-l[1]) / norm(l[0]-l[1]);
	return l[0] + t*(l[0]-l[1]);
}

point reflection(const line& l, const point& p) {
	return p + 2. * (projection(l, p) - p);
}
 typedef array<point, 2> line;
//using line = array<point, 2>;
/*
 * 平行かどうか
 */
bool is_parallel(const line& s, const line& t) {
	return eq(cross(s[0] - s[1], t[0] - t[1]), 0.0);
}

bool is_orthogonal(const line& s, const line& t) {
	return eq(dot(s[0] - s[1], t[0] - t[1]), 0.0);
}

bool intersectLL(const line& l, const line& m) {
	return !eq(cross(l[1]-l[0], m[1]-m[0]), 0.) || // non-parallel
			eq(cross(l[1]-l[0], m[0]-l[0]), 0.);   // same line
}
bool intersectSS(const line &s, const line &t) {
	return ccw(s[0],s[1],t[0])*ccw(s[0],s[1],t[1]) <= 0 &&
			ccw(t[0],t[1],s[0])*ccw(t[0],t[1],s[1]) <= 0;
}

point crosspoint(const line& l, const line& m) {
	double A = cross(l[1] - l[0], m[1] - m[0]);
	double B = cross(l[1] - l[0], l[1] - m[0]);
	if (eq(A, 0.) && eq(B, 0.)) return m[0]; // same line
	if (eq(A, 0.)) assert(false); // 交点が無い場合に落ちる 実行前に交差判定
	return m[0] + B / A * (m[1] - m[0]);
}

int main()
{
  int q;
  cin>>q;
  for(;q--;)
    {
      line l1,l2;
      double x,y;
      cin>>x>>y;
      l1[0]=point(x,y);
      cin>>x>>y;
      l1[1]=point(x,y);
      cin>>x>>y;
      l2[0]=point(x,y);
      cin>>x>>y;
      l2[1]=point(x,y);

      if(intersectSS(l1,l2))
	{
	  point p=crosspoint(l1,l2);
	  printf("%.10lf %.10lf\n",p.real(),p.imag());
	  //	  cout<< p.real()<<" "intersectSS(l1,l2) <<endl;
	}
    }
}
### Prompt

Your challenge is to write a Cpp solution to the following 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

### Response

```cpp
#include<iostream>
#include<complex>
#include<array>
#include<vector>
#include<cstdio>
#include<cassert>
using namespace std;

const double EPS = 1e-10;
template<class T>
bool eq(const T& a, const T& b) {return abs(a - b) < EPS;}

typedef  complex<double> point;

double cross(const point& a, const point& b) {
	return imag(conj(a)*b);
}
double dot(const point& a, const point& b) {
	return real(conj(a)*b);
}
int ccw(point a, point b, point c) {
	b -= a; c -= a;
	if (cross(b, c) > 0)   return +1;  // counter clockwise
	if (cross(b, c) < 0)   return -1;  // clockwise
	if (dot(b, c) < 0)     return +2;  // c--a--b on line
	if (norm(b) < norm(c)) return -2;  // a--b--c on line
	return 0;  // a--c--b on line or a----bc
}

typedef array<point, 2> line;

point projection(const line& l, const point& p) {
	double t = dot(p-l[0], l[0]-l[1]) / norm(l[0]-l[1]);
	return l[0] + t*(l[0]-l[1]);
}

point reflection(const line& l, const point& p) {
	return p + 2. * (projection(l, p) - p);
}
 typedef array<point, 2> line;
//using line = array<point, 2>;
/*
 * 平行かどうか
 */
bool is_parallel(const line& s, const line& t) {
	return eq(cross(s[0] - s[1], t[0] - t[1]), 0.0);
}

bool is_orthogonal(const line& s, const line& t) {
	return eq(dot(s[0] - s[1], t[0] - t[1]), 0.0);
}

bool intersectLL(const line& l, const line& m) {
	return !eq(cross(l[1]-l[0], m[1]-m[0]), 0.) || // non-parallel
			eq(cross(l[1]-l[0], m[0]-l[0]), 0.);   // same line
}
bool intersectSS(const line &s, const line &t) {
	return ccw(s[0],s[1],t[0])*ccw(s[0],s[1],t[1]) <= 0 &&
			ccw(t[0],t[1],s[0])*ccw(t[0],t[1],s[1]) <= 0;
}

point crosspoint(const line& l, const line& m) {
	double A = cross(l[1] - l[0], m[1] - m[0]);
	double B = cross(l[1] - l[0], l[1] - m[0]);
	if (eq(A, 0.) && eq(B, 0.)) return m[0]; // same line
	if (eq(A, 0.)) assert(false); // 交点が無い場合に落ちる 実行前に交差判定
	return m[0] + B / A * (m[1] - m[0]);
}

int main()
{
  int q;
  cin>>q;
  for(;q--;)
    {
      line l1,l2;
      double x,y;
      cin>>x>>y;
      l1[0]=point(x,y);
      cin>>x>>y;
      l1[1]=point(x,y);
      cin>>x>>y;
      l2[0]=point(x,y);
      cin>>x>>y;
      l2[1]=point(x,y);

      if(intersectSS(l1,l2))
	{
	  point p=crosspoint(l1,l2);
	  printf("%.10lf %.10lf\n",p.real(),p.imag());
	  //	  cout<< p.real()<<" "intersectSS(l1,l2) <<endl;
	}
    }
}
```