#include<iostream>
#include<cmath>
#include<iomanip>
using namespace std; long double X1[100000], X2[100000], X3[100000], X4[100000], Y1[100000], Y2[100000], Y3[100000], Y4[100000], INF = 1e+14, EPS = 1e-14, Q;
bool paralel(long double X1, long double Y1, long double X2, long double Y2, long double X3, long double Y3, long double X4, long double Y4) {
	long double xp[4], yp[4]; xp[0] = X1 - X2; yp[0] = Y1 - Y2; xp[1] = X3 - X4; yp[1] = Y3 - Y4;
	if (yp[0] == 0 && yp[1] == 0) { return true; }if (yp[0] == 0 && yp[1] != 0) { return false; }
	if (yp[0] != 0 && yp[1] == 0) { return false; }if ((xp[0] / yp[0]) == (xp[1] / yp[1])) { return true; }return false;
}
pair<long double, long double>kouten(long double X1, long double Y1, long double X2, long double Y2, long double X3, long double Y3, long double X4, long double Y4) {
	if (paralel(X1, Y1, X2, Y2, X3, Y3, X4, Y4) == true) { return make_pair(INF, INF); }
	if (fabs(X1 - X2) < EPS) { X1 -= EPS; }if (fabs(X3 - X4) < EPS) { X3 -= EPS; }
	if (fabs(Y1 - Y2) < EPS) { Y1 -= EPS; }if (fabs(Y3 - Y4) < EPS) { Y3 -= EPS; }long double XP[4], YP[4], ZP[4];
	XP[0] = X1 - X2; YP[0] = Y1 - Y2; XP[1] = X3 - X4; YP[0] = Y1 - Y2; YP[1] = Y3 - Y4; ZP[0] = YP[0] / XP[0]; ZP[1] = YP[1] / XP[1];
	long double X5 = X1, Y5 = Y3 - (X3 - X1)*ZP[1], K = (Y5 - Y1) / (ZP[0] - ZP[1]); long double L = K*ZP[0]; return make_pair(X1 + K, Y1 + L);
}
void solve(long double X1, long double Y1, long double X2, long double Y2, long double X3, long double Y3, long double X4, long double Y4) { 
	pair<long double, long double>pa = kouten(X1, Y1, X2, Y2, X3, Y3, X4, Y4); 
	cout << fixed << setprecision(15) << pa.first << ' ' << fixed << setprecision(15) << pa.second << endl; 
} int main() { cin >> Q; for (int i = 0; i < Q; i++) { cin >> X1[i] >> Y1[i] >> X2[i] >> Y2[i] >> X3[i] >> Y3[i] >> X4[i] >> Y4[i];; solve(X1[i], Y1[i], X2[i], Y2[i], X3[i], Y3[i], X4[i], Y4[i]); } return 0; }
### 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<cmath>
#include<iomanip>
using namespace std; long double X1[100000], X2[100000], X3[100000], X4[100000], Y1[100000], Y2[100000], Y3[100000], Y4[100000], INF = 1e+14, EPS = 1e-14, Q;
bool paralel(long double X1, long double Y1, long double X2, long double Y2, long double X3, long double Y3, long double X4, long double Y4) {
	long double xp[4], yp[4]; xp[0] = X1 - X2; yp[0] = Y1 - Y2; xp[1] = X3 - X4; yp[1] = Y3 - Y4;
	if (yp[0] == 0 && yp[1] == 0) { return true; }if (yp[0] == 0 && yp[1] != 0) { return false; }
	if (yp[0] != 0 && yp[1] == 0) { return false; }if ((xp[0] / yp[0]) == (xp[1] / yp[1])) { return true; }return false;
}
pair<long double, long double>kouten(long double X1, long double Y1, long double X2, long double Y2, long double X3, long double Y3, long double X4, long double Y4) {
	if (paralel(X1, Y1, X2, Y2, X3, Y3, X4, Y4) == true) { return make_pair(INF, INF); }
	if (fabs(X1 - X2) < EPS) { X1 -= EPS; }if (fabs(X3 - X4) < EPS) { X3 -= EPS; }
	if (fabs(Y1 - Y2) < EPS) { Y1 -= EPS; }if (fabs(Y3 - Y4) < EPS) { Y3 -= EPS; }long double XP[4], YP[4], ZP[4];
	XP[0] = X1 - X2; YP[0] = Y1 - Y2; XP[1] = X3 - X4; YP[0] = Y1 - Y2; YP[1] = Y3 - Y4; ZP[0] = YP[0] / XP[0]; ZP[1] = YP[1] / XP[1];
	long double X5 = X1, Y5 = Y3 - (X3 - X1)*ZP[1], K = (Y5 - Y1) / (ZP[0] - ZP[1]); long double L = K*ZP[0]; return make_pair(X1 + K, Y1 + L);
}
void solve(long double X1, long double Y1, long double X2, long double Y2, long double X3, long double Y3, long double X4, long double Y4) { 
	pair<long double, long double>pa = kouten(X1, Y1, X2, Y2, X3, Y3, X4, Y4); 
	cout << fixed << setprecision(15) << pa.first << ' ' << fixed << setprecision(15) << pa.second << endl; 
} int main() { cin >> Q; for (int i = 0; i < Q; i++) { cin >> X1[i] >> Y1[i] >> X2[i] >> Y2[i] >> X3[i] >> Y3[i] >> X4[i] >> Y4[i];; solve(X1[i], Y1[i], X2[i], Y2[i], X3[i], Y3[i], X4[i], Y4[i]); } return 0; }
```