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
#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;

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 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);
        if (is_parallel(p0,p1,p2,p3)) cout << 2 << endl;
        else if (is_orthogonal(p0,p1,p2,p3)) cout << 1 << endl;
        else cout << 0 << endl;
    }
    return 0;
}
