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 <iostream>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
using namespace std;
int num;
double x,yy1,yy2,yy3,yy4;
const double eps = 1e-8;
const double inf = 1e20;
const double pi = acos(-1.0);
int sgn(double x){
    if(fabs(x)<eps){
        return 0;
    }
    if(x<0){
        return -1;
    }else{
        return 1;
    }
}
struct Point{
    double x,y;
    Point(){}
    void input(){
        scanf("%lf%lf",&x,&y);
    }
    Point(double _x,double _y){
        x = _x;y = _y;
    }
    bool operator < (Point b)const{
        return sgn(x-b.x)==0?sgn(y-b.y)<0:x<b.x;
    }
    Point operator - (const Point &b)const{
        return Point(x-b.x,y-b.y);
    }
    double operator *(const Point &b)const{
        return x*b.x+y*b.y;
    }
    double operator ^(const Point &b)const{
        return x*b.y-y*b.x;
    }
    double distance(Point p){
        return hypot(x-p.x,y-p.y);
    }
};
struct Line{
    Point s,e;
    Line(){}
    Line(Point _s,Point _e){
        s = _s;e = _e;
    }
    //求两直线的交点
    // 要保证两直线不平行或重合
    Point crosspoint(Line v){
        double a1 = (v.e-v.s)^(s-v.s);
        double a2 = (v.e-v.s)^(e-v.s);
        return Point((s.x*a2-e.x*a1)/(a2-a1),(s.y*a2-e.y*a1)/(a2-a1));
    }
};
int main(){
    int q;
    scanf("%d",&q);
    Point p0,p1,p2,p3;
    while(q--){
        p0.input();
        p1.input();
        p2.input();
        p3.input();
        Line line1(p0,p1);
        Line line2(p2,p3);
        Point ans = line1.crosspoint(line2);
        printf("%.10f %.10f\n",ans.x,ans.y);
    }
   
    return 0;
}