#include <bits/stdc++.h>
using namespace std;
typedef complex<double> Point;
typedef pair<Point, Point> Line;
const double EPS = 1e-10;

// 内積 |a||b|cosθ
double dot(Point a, Point b){
	// return a.real() * b.real() + a.imag() * b.imag();
	return (conj(a) * b).real();
}
// ベクトルa, bがなす角の余弦(cosθ)の値
double costheta(Point a, Point b){
    return dot(a, b) / (abs(a) * abs(b));
}
// 2直線の平行判定
bool isParallel(Line a, Line b){
    double cos = costheta(a.second - a.first, b.second - b.first);
    return fabs(cos) > 1 - EPS;
}
// 2直線が垂直に交わるかどうか判定
bool isOrthogonal(Line a, Line b){
    double cos = costheta(a.second - a.first, b.second - b.first);
    return fabs(cos) < EPS;
}

int main(){
    int q;
    cin >> q;
    for(int i = 0; i < q; i++){
        double x, y;
        Point p[4];
        for(int j = 0; j < 4; j++){
            cin >> x >> y;
            p[j] = {x, y};
        }
        Line s1 = {p[0], p[1]};
        Line s2 = {p[2], p[3]};
        if(isOrthogonal(s1, s2)) cout << 1 << endl;
        else if(isParallel(s1, s2)) cout << 2 << endl;
        else cout << 0 << endl;
    }
}

### Prompt

Develop a solution in cpp to the problem described below:

For given two lines s1 and s2, print "2" if they are parallel, "1" if they are orthogonal, or "0" otherwise.

s1 crosses points p0 and p1, and s2 crosses points p2 and p3.

Constraints

* 1 ≤ q ≤ 1000
* -10000 ≤ xpi, ypi ≤ 10000
* p0 ≠ p1 and p2 ≠ p3.

Input

The entire input looks like:


q (the number of queries)
1st query
2nd query
...
qth query


Each query consists of integer coordinates of the points p0, p1, p2, p3 in the following format:


xp0 yp0 xp1 yp1 xp2 yp2 xp3 yp3


Output

For each query, print "2", "1" or "0".

Example

Input

3
0 0 3 0 0 2 3 2
0 0 3 0 1 1 1 4
0 0 3 0 1 1 2 2


Output

2
1
0

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
typedef complex<double> Point;
typedef pair<Point, Point> Line;
const double EPS = 1e-10;

// 内積 |a||b|cosθ
double dot(Point a, Point b){
	// return a.real() * b.real() + a.imag() * b.imag();
	return (conj(a) * b).real();
}
// ベクトルa, bがなす角の余弦(cosθ)の値
double costheta(Point a, Point b){
    return dot(a, b) / (abs(a) * abs(b));
}
// 2直線の平行判定
bool isParallel(Line a, Line b){
    double cos = costheta(a.second - a.first, b.second - b.first);
    return fabs(cos) > 1 - EPS;
}
// 2直線が垂直に交わるかどうか判定
bool isOrthogonal(Line a, Line b){
    double cos = costheta(a.second - a.first, b.second - b.first);
    return fabs(cos) < EPS;
}

int main(){
    int q;
    cin >> q;
    for(int i = 0; i < q; i++){
        double x, y;
        Point p[4];
        for(int j = 0; j < 4; j++){
            cin >> x >> y;
            p[j] = {x, y};
        }
        Line s1 = {p[0], p[1]};
        Line s2 = {p[2], p[3]};
        if(isOrthogonal(s1, s2)) cout << 1 << endl;
        else if(isParallel(s1, s2)) cout << 2 << endl;
        else cout << 0 << endl;
    }
}

```