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

#define dump(n) cout<<"# "<<#n<<'='<<(n)<<endl
#define repi(i,a,b) for(int i=int(a);i<int(b);i++)
#define peri(i,a,b) for(int i=int(b);i-->int(a);)
#define rep(i,n) repi(i,0,n)
#define per(i,n) peri(i,0,n)
#define all(c) begin(c),end(c)
#define mp make_pair
#define mt make_tuple

typedef unsigned int uint;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<string> vs;

const int INF=1e9;
const int MOD=1e9+7;
const double EPS=1e-9;

template<typename T1,typename T2>
ostream& operator<<(ostream& os,const pair<T1,T2>& p){
	return os<<'('<<p.first<<','<<p.second<<')';
}
template<typename T>
ostream& operator<<(ostream& os,const vector<T>& a){
	os<<'[';
	rep(i,a.size()) os<<(i?" ":"")<<a[i];
	return os<<']';
}

struct Point{
	double x,y;
	Point(){}
	Point(double x,double y):x(x),y(y){}
	Point& operator+=(Point p){
		x+=p.x,y+=p.y;
		return *this;
	}
	Point& operator-=(Point p){
		x-=p.x,y-=p.y;
		return *this;
	}
	Point& operator*=(double c){
		x*=c,y*=c;
		return *this;
	}
	Point& operator/=(double c){
		x/=c,y/=c;
		return *this;
	}
};
Point operator+(Point a,Point b){
	return a+=b;
}
Point operator-(Point a,Point b){
	return a-=b;
}
Point operator*(Point a,double c){
	return a*=c;
}
Point operator*(double c,Point a){
	return a*=c;
}
Point operator/(Point a,double c){
	return a/=c;
}
bool operator==(Point a,Point b){
	return abs(a.x-b.x)<EPS && abs(a.y-b.y)<EPS;
}
bool operator!=(Point a,Point b){
	return !(a==b);
}

double Abs(Point p){
	return sqrt(p.x*p.x+p.y*p.y);
}
double Abs2(Point p){
	return p.x*p.x+p.y*p.y;
}
double Arg(Point p){
	return atan2(p.y,p.x);
}
double Dot(Point a,Point b){
	return a.x*b.x+a.y*b.y;
}
double Cross(Point a,Point b){
	return a.x*b.y-a.y*b.x;
}
Point Rot(Point p,double t){
	return Point(cos(t)*p.x-sin(t)*p.y,sin(t)*p.x+cos(t)*p.y);
}

struct Line{
	Point pos,dir;
	Line(){}
	Line(Point p,Point d):pos(p),dir(d){}
	Line(double px,double py,double dx,double dy):pos(px,py),dir(dx,dy){}
};

// aとbは必ず交差していること
// 同一直線上にある場合，a.posを返す
Point InterPointLL(Line a,Line b){
	if(abs(Cross(a.dir,b.dir))<EPS) return a.pos;
	return a.pos+Cross(b.pos-a.pos,b.dir)/Cross(a.dir,b.dir)*a.dir;
}

void solve()
{
	Point a,b,c,d;
	cin>>a.x>>a.y>>b.x>>b.y>>c.x>>c.y>>d.x>>d.y;
	Point p=InterPointLL(Line(a,b-a),Line(c,d-c));
	printf("%.10f %.10f\n",p.x,p.y);
}

int main()
{
	int q; cin>>q;
	rep(_,q) solve();
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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 <bits/stdc++.h>
using namespace std;

#define dump(n) cout<<"# "<<#n<<'='<<(n)<<endl
#define repi(i,a,b) for(int i=int(a);i<int(b);i++)
#define peri(i,a,b) for(int i=int(b);i-->int(a);)
#define rep(i,n) repi(i,0,n)
#define per(i,n) peri(i,0,n)
#define all(c) begin(c),end(c)
#define mp make_pair
#define mt make_tuple

typedef unsigned int uint;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<string> vs;

const int INF=1e9;
const int MOD=1e9+7;
const double EPS=1e-9;

template<typename T1,typename T2>
ostream& operator<<(ostream& os,const pair<T1,T2>& p){
	return os<<'('<<p.first<<','<<p.second<<')';
}
template<typename T>
ostream& operator<<(ostream& os,const vector<T>& a){
	os<<'[';
	rep(i,a.size()) os<<(i?" ":"")<<a[i];
	return os<<']';
}

struct Point{
	double x,y;
	Point(){}
	Point(double x,double y):x(x),y(y){}
	Point& operator+=(Point p){
		x+=p.x,y+=p.y;
		return *this;
	}
	Point& operator-=(Point p){
		x-=p.x,y-=p.y;
		return *this;
	}
	Point& operator*=(double c){
		x*=c,y*=c;
		return *this;
	}
	Point& operator/=(double c){
		x/=c,y/=c;
		return *this;
	}
};
Point operator+(Point a,Point b){
	return a+=b;
}
Point operator-(Point a,Point b){
	return a-=b;
}
Point operator*(Point a,double c){
	return a*=c;
}
Point operator*(double c,Point a){
	return a*=c;
}
Point operator/(Point a,double c){
	return a/=c;
}
bool operator==(Point a,Point b){
	return abs(a.x-b.x)<EPS && abs(a.y-b.y)<EPS;
}
bool operator!=(Point a,Point b){
	return !(a==b);
}

double Abs(Point p){
	return sqrt(p.x*p.x+p.y*p.y);
}
double Abs2(Point p){
	return p.x*p.x+p.y*p.y;
}
double Arg(Point p){
	return atan2(p.y,p.x);
}
double Dot(Point a,Point b){
	return a.x*b.x+a.y*b.y;
}
double Cross(Point a,Point b){
	return a.x*b.y-a.y*b.x;
}
Point Rot(Point p,double t){
	return Point(cos(t)*p.x-sin(t)*p.y,sin(t)*p.x+cos(t)*p.y);
}

struct Line{
	Point pos,dir;
	Line(){}
	Line(Point p,Point d):pos(p),dir(d){}
	Line(double px,double py,double dx,double dy):pos(px,py),dir(dx,dy){}
};

// aとbは必ず交差していること
// 同一直線上にある場合，a.posを返す
Point InterPointLL(Line a,Line b){
	if(abs(Cross(a.dir,b.dir))<EPS) return a.pos;
	return a.pos+Cross(b.pos-a.pos,b.dir)/Cross(a.dir,b.dir)*a.dir;
}

void solve()
{
	Point a,b,c,d;
	cin>>a.x>>a.y>>b.x>>b.y>>c.x>>c.y>>d.x>>d.y;
	Point p=InterPointLL(Line(a,b-a),Line(c,d-c));
	printf("%.10f %.10f\n",p.x,p.y);
}

int main()
{
	int q; cin>>q;
	rep(_,q) solve();
}
```