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 <cstdio>
#include <utility>
#include <complex>
#include <array>
using namespace std;

#ifdef __linux
#define _U(s) s##_unlocked
#else
#define _U(s) _##s##_nolock
#define _CRT_DISABLE_PERFCRIT_LOCKS
#endif
#define gcu _U(getchar)
#define pcu _U(putchar)
#define _DEF(r, n, ...) inline r n(__VA_ARGS__) noexcept
#define _T template <typename T>
#define _HT template <typename H,typename... T>
#define _OP(t) inline operator t()
struct _in {
#ifdef _GLIBCXX_STRING
	_OP(string){string s;for(char c;c=gcu(),c!=' '&&c!='\n';)s+=c;return s;}
#define _S
#endif
	_OP(char){char c=gcu();gcu();return c;}
	_OP(double){double d; scanf("%lf",&d); gcu();return d;}
	_T _OP(T){T n{},m{1},c;if((c=gcu())=='-')m=-1,c=gcu();do{n=10*n+(c-'0'),c=gcu();}while(c>='0'&&c<='9');return m*n;}
} in;
#define _SCAN(...) _DEF(bool,scan,__VA_ARGS__)
#ifdef _S
_SCAN(string &o) {int c{gcu()};if(c==EOF)return false;else{ungetc(c,stdin);string t=move(in);o=t;return true;}}
#endif
_T _SCAN(T &o) {int c{gcu()};return c==EOF?false:(ungetc(c,stdin),o=in,true);}
_HT _SCAN(H &h,T&&... t){return scan(h)&&scan(t...);}
#define _OUT(...) _DEF(void,out,__VA_ARGS__)
#define _OUTL(...) _DEF(void,outl,__VA_ARGS__)
_OUT(bool b){pcu('0'+b);}
_OUT(const char *s){while(*s)pcu(*s++);}
_OUT(char c){pcu(c);}
#ifdef _S
_OUT(string &s){for(char c:s)pcu(c);}
#endif
_T _OUT(T n){static char b[20];char *p=b;T m=n<0?pcu('-'),-1:1;
	if(!n)*p++='0';else while(n)*p++=(char)(n%10*m+'0'),n/=10;while(p!=b)pcu(*--p);}
_OUTL(){out('\n');}
#ifdef _GLIBCXX_VECTOR
	_T _OUT(vector<T> &v){for(T &x:v)out(&x == &v[0]?"":" "),out(x);}
#endif
	_HT _OUT(H &&h, T... t){out(h);out(t...);}
	template <typename... T> _OUTL(T... t){out(t...);outl();}
	struct range{
		int e,b=0,s=1;range(int _b,int _e,int _s):e(_e),b(_b),s(_s){} range(int _b,int _e): e(_e), b(_b){} range(int _e):e(_e){}
		struct it{int v,s; it(int _v,int _s):v(_v),s(_s){} operator int()const{return v;} operator int&(){return v;}int operator*()const{return v;}
			it& operator++(){v+=s;return *this;} }; it begin(){return {b,s};} it end(){return {e,s};}};
#define times(i,n) for(int i=n;i;i--)

using P = complex<double>;
void dump(P p) {printf("%.10f %.10f\n", real(p), imag(p));}

int main() {
	times(i, in) {
		array<P, 4> p;
		for (P &i: p) {
			double x = (int) in, y = (int) in;
			i = {x, y};
		}
		P b = p[3] - p[2];
		auto D = [&](P &a) {return abs(imag(b *  conj(a - p[2])));};
		double d1 = D(p[0]);
		dump(p[0] + (p[1] - p[0]) * d1 / (d1 + D(p[1])));
	}
}