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>
#include <complex>
#define rep(i, n) for (int i = 0; i < (n); i++)
#define loop(i, x, n) for (int i = (x); i < (n); i++)
#define all(v) (v).begin(), (v).end()
#define EQ(a, b) (abs((a) - (b)) < EPS)
#define int long long
#define INF 1e9
#define MOD 1e9 + 7
using namespace std;

template<typename T> void cmin(T &a, T b) { a = min(a, b); }
template<typename T> void cmax(T &a, T b) { a = max(a, b); }

using P = complex<double>;
const double PI = acos(-1);
const double EPS = 1e-9;
// P p(a,b)=a+bi -> 点p(a,b)
namespace std {
  bool operator<(const P &a, const P &b) { return real(a) != real(b) ? real(a) < real(b) : imag(a) < imag(b); }
  bool operator==(const P &a, const P &b) { return a.real() == b.real() && a.imag() == b.imag(); }
} // namespace std

struct L : public vector<P> {
  L(const P &a, const P &b) {
    push_back(a);
    push_back(b);
  }
};
//内積
double dot(P a, P b) { return real(conj(a) * b); }
//外戚
double cross(P a, P b) { return imag(conj(a) * b); }

P crosspointSS(L a, L b) {
  double t1 = abs(cross(a[1] - a[0], b[0] - a[0]));
  double t2 = abs(cross(a[1] - a[0], b[1] - a[0]));
  return b[0] + (b[1] - b[0]) * t1 / (t1 + t2);
}

signed main() {
  int q;
  cin >> q;
  while (q--) {
    int a, b, c, d, e, f, g, h;
    cin >> a >> b >> c >> d >> e >> f >> g >> h;
    L l(P(a, b), P(c, d)), m(P(e, f), P(g, h));
    P ans = crosspointSS(l, m);
    printf("%.10lf %.10lf\n", ans.real(), ans.imag());
  }
  return 0;
}