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>
using namespace std;
const double eps=1e-9;
struct dot{
	double x,y;
	double len(){return sqrt(x*x+y*y);}
	dot(double a=0.0,double b=0.0){x=a,y=b;}
	dot operator +(const dot&b)const {return dot(x+b.x,y+b.y);}
	dot operator -(const dot&b)const {return dot(x-b.x,y-b.y);}
	dot operator *(const double&z)const {return dot(x*z,y*z);}
	bool operator <(const dot&b)const {return (x<b.x)||(x==b.x&&y<b.y);}
	void read(){scanf("%lf%lf",&x,&y);} 
};
struct line{
	dot x1,x2;
	void read(){x1.read();x2.read();};
	dot getdot(){return x2-x1;}
};
double chaji(dot x,dot y){
	return x.x*y.y-x.y*y.x;
}
double neiji(dot x,dot y){
	return x.x*y.x+x.y*y.y;
}
double cos(dot x,dot y){
	if (!x.len()||!y.len())return 0;
	return neiji(x,y)/x.len()/y.len();
}
dot Projection(dot x,line y){//求x关于直线y的投影点（垂足） 
	y.x2=y.x2-y.x1;
	x=x-y.x1;
	double len=cos(x,y.x2)*x.len()/y.x2.len();
	y.x2=y.x2*len;
	return y.x2+y.x1;
}
dot Reflection(dot x,line y){//求x关于直线y的对称点 
	dot z=Projection(x,y);
	return z*2-x;
}
int Counter_Clockwise(dot x,line y){//求两个向量的五种关系 
	x=x-y.x1;y.x2=y.x2-y.x1;
	if (chaji(y.x2,x)!=0){
		if (chaji(y.x2,x)>0)return 1;
		return 2;
	} 
	if (fabs(cos(x,y.x2)+1)<eps)return 3;
	if (x.len()>y.x2.len())return 4;
	return 5;
} 
int Parallel_Orthogonal(line x,line y){//求两个直线是垂直还是平行还是其他 
	dot a=x.getdot(),b=y.getdot();
	if (chaji(a,b)==0)return 2;
	if (neiji(a,b)==0)return 1;
	return 0;
}
int main(){
	int T;
	scanf("%d",&T);
	while (T--){
		line x,y;
		x.read();y.read();
		printf("%d\n",Parallel_Orthogonal(x,y));
	}
	return 0;
}

