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>
#define EPS 1e-10
#define equals(a,b) (fabs((a)-(b)) < EPS)
#define rep(i,n) for(int i=0;i<n;++i)
typedef long long ll;
using namespace std;

static const int COUNTER_CLOCKWISE = 1;
static const int CLOCKWISE = -1;
static const int ONLINE_BACK = 2;
static const int ONLINE_FRONT = -2;
static const int ON_SEGMENT = 0;

struct Point {
    double x,y;
    Point(){}
    Point(double x, double y) : x(x),y(y){}
    Point operator+(Point p) {return Point(x+p.x, y+p.y);}
    Point operator-(Point p) {return Point(x-p.x, y-p.y);}
    Point operator*(double k){return Point(x*k,y*k);}
    Point operator/(double k){return Point(x/k,y/k);}
    double norm(){return x*x+y*y;}
    double abs(){sqrt(norm());}
    bool operator == (const Point &p) const {return equals(x,p.x)&&equals(y,p.y);}
};
typedef Point P;

double norm(P a){return a.x*a.x+a.y*a.y;}
double abs(P a){return sqrt(norm(a));}

// 線分
struct Segment {
    Point p1,p2;
    Segment(){}
    Segment(Point p1, Point p2):p1(p1),p2(p2){}
};

// 内積
double dot(Point a, Point b) {return a.x*b.x + a.y*b.y;}

// 外積
double cross(Point a, Point b) {return a.x*b.y-a.y*b.x;}
// 2直線の直行判定
bool is_orthogonal(Point a1, Point a2, Point b1, Point b2) {
    return equals(dot(a1-a2, b1-b2), 0.0);
}
// 2直線の平行判定
bool is_parallel(Point a1, Point a2, Point b1, Point b2) {
    return equals(cross(a1-a2, b1-b2), 0.0);
}

int ccw(Point p0,Point p1,Point p2) {
    P a = p1-p0;
    P b = p2-p0;
    if(cross(a,b) > EPS) return COUNTER_CLOCKWISE;
    if(cross(a,b) < -EPS) return CLOCKWISE;
    if(dot(a,b) < -EPS) return ONLINE_BACK;
    if(a.norm()<b.norm()) return ONLINE_FRONT;
    return ON_SEGMENT;
}

bool is_intersected_ls(Point p1, Point p2, Point p3, Point p4) {
    return (ccw(p1,p2,p3)*ccw(p1,p2,p4) <= 0&&ccw(p3,p4,p1)*ccw(p3,p4,p2) <= 0 );
}

P intersection_ls(P a1, P a2, P b1, P b2) {
    P b = b2-b1;
    double d1 = abs(cross(b, a1-b1));
    double d2 = abs(cross(b, a2-b1));
    double t = d1/(d1+d2);
    return a1+(a2-a1)*t;
}

int main() {
    int q;
    cin>>q;
    rep(i,q){
        double x0,y0,x1,y1,x2,y2,x3,y3;
        cin>>x0>>y0>>x1>>y1>>x2>>y2>>x3>>y3;
        P p0(x0,y0),p1(x1,y1),p2(x2,y2),p3(x3,y3);
        P ans = intersection_ls(p0,p1,p2,p3);
        cout << fixed << setprecision(12);
        cout << ans.x << " " << ans.y << endl;
    }
    return 0;
}