Problem: For given a circle $c$ and a line $l$, print the coordinates of the cross points of them.

Constraints

* $p1$ and $p2$ are different
* The circle and line have at least one cross point
* $1 \leq q \leq 1,000$
* $-10,000 \leq cx, cy, x1, y1, x2, y2 \leq 10,000$
* $1 \leq r \leq 10,000$

Input

The input is given in the following format.

$cx\; cy\; r$
$q$
$Line_1$
$Line_2$
:
$Line_q$


In the first line, the center coordinate of the circle and its radius are given by $cx$, $cy$ and $r$. In the second line, the number of queries $q$ is given.

In the following $q$ lines, as queries, $Line_i$ are given ($1 \leq i \leq q$) in the following format.

$x_1\; y_1\; x_2\; y_2$

Each line is represented by two points $p1$ and $p2$ which the line crosses. The coordinate of $p1$ and $p2$ are given by ($x1$, $y1$) and ($x2$, $y2$) respectively. All input values are given in integers.

Output

For each query, print the coordinates of the cross points in the following rules.

* If there is one cross point, print two coordinates with the same values.
* Print the coordinate with smaller $x$ first. In case of a tie, print the coordinate with smaller $y$ first.



The output values should be in a decimal fraction with an error less than 0.000001.

Example

Input

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


Output

1.00000000 1.00000000 3.00000000 1.00000000
3.00000000 1.00000000 3.00000000 1.00000000

Solution (in CPP):

Solution:
//计算几何 板子
#include<bits/stdc++.h>
using namespace std ;
#define eps 1e-10
#define equals(a , b) (fabs((a) - (b)) < eps)

static const int COUNTER_CLOCKWISE = 1;
static const int CLOCKWISE = -1 ;
static const int ONCLICK_BACK = 2 ;
static const int ONLINE_FRONT = -2 ;
static const int ON_SEGMENT = 0 ;

class Point
{
public:
    double x , y;

    Point(double x = 0 ,double y = 0):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 a){return Point(a * x , a * y) ;}
    Point operator /(double a){return Point(x / a , y / a) ;}

    double abs() {return sqrt(norm()) ;}
    double norm() {return x * x + y * y ;}

    bool operator < (const Point &p) const
    {
        return x != p.x ? x < p.x : y < p.y ;
    }

    bool operator == (const Point &p) const
    {
        return fabs(x - p.x) < eps && fabs(y - p.y) < eps ;
    }

};

typedef Point Vector ;

struct Segment  //表示线段结构体
{
	Point p1,p2;
	Segment(){}
	Segment(Point _p1,Point _p2):p1(_p1),p2(_p2){}
};

typedef Segment Line ;

class Circle{  //表示圆的结构体
public:
    Point c ;
    double r ;
    Circle(Point c = Point() , double r = 0.0):c(c) ,r(r){}
};

double dot(Vector a , Vector b)  // 向量a,b的内积,就是向量a*b
{
    return a.x * b.x + a.y * b.y ;
}

double cross(Vector a, Vector b)  //向量a,b外积
{
	return a.x * b.y - a.y * b.x ;
}

bool isOrthogonal(Vector a , Vector b)  //判断是否正交
{
    return equals(dot(a , b) , 0.0) ;
}

bool isOrthogonal(Point a1 , Point a2 , Point b1 , Point b2)
{
    return isOrthogonal(a1 - a2 , b1 - b2) ;
}

bool  isOrthogonal(Segment s1 , Segment s2)
{
    return equals(dot(s1.p2 - s1.p1 , s2.p2 - s2.p1) , 0.0) ;
}

bool isParallel(Vector a, Vector b)  //判断是否平行
{
	return equals(cross(a, b), 0.0) ;
}

bool isParallel(Point a1, Point a2, Point b1, Point b2)
{
	return isParallel(a1 - a2, b1 - b2) ;
}

bool isParallel(Segment s1, Segment s2)
{
	return equals(cross(s1.p2 - s1.p1, s2.p2 - s2.p1), 0.0) ;
}

Point project(Segment s, Point p)  //p在s上的投影
{
    Point base = s.p2 - s.p1 ;
    double r = dot(p - s.p1 , base) / base.norm() ;
    return s.p1 +base * r ;
}
/*
求垂足x：对于给定的三点p1,p2,p从点p向通过p1,p2的直线引一条垂线
base=p2-p1;
hypo=p-p1;
 x=s.p1+base*(hypo*base/|base|^2); hypo*base 可以用向量内积求
*/

Point reflect(Segment s , Point p)
{
    return p + (project(s , p) - p) * 2.0 ;
}

int ccw(Point p0 , Point p1 , Point p2)//判断三个点相对位置  // 判断逆时针方向的时候也可以使用
{
	Point a = p1 - p0 ;
	Point b = p2 - p0 ;
	if(cross(a , b) > eps) return 1 ;  //p0,p1,p2成逆时针方向
	if(cross(a , b) < -eps) return -1 ;  //p0,p1,p2成顺时针方向
	if(dot(a , b) < -eps) return 2 ;  //p2 p0 p1一次排列在同一直线上
	if(a.norm() < b.norm()) return -2 ;  // p0 p1 p2一次排列在同一直线上
	return 0 ;  //p2在线段p0p1上
}

bool intersect(Point p1,Point p2,Point p3,Point p4)  //判断是否相交 通过ccw来判断
{
	return (ccw(p1,p2,p3)*ccw(p1,p2,p4)<=0&&ccw(p3,p4,p1)*ccw(p3,p4,p2)<=0);
}

bool intersect(Segment s1 , Segment s2)
{
    return intersect(s1.p1 , s1.p2 , s2.p1 , s2.p2) ;
}

double getdis(Point a , Point b)  //求俩点间的距离
{
	return (a - b) . abs() ; //点a与点b之间的距离等于向量a-b或b-a的绝对值。
}

//注意：这里Segmet s是不规范的应该是 Line s 才对
double getdis_Li(Segment s , Point p)  //点与直线的距离
{
	return fabs(cross(s.p2 - s.p1 , p - s.p1)/(s.p2-s.p1).abs());
}

double getdis_sp(Segment s , Point p)  //点与线段的距离
{
	if(dot(s.p2 - s.p1 , p - s.p1) < 0.0) return (p-s.p1).abs();
	if(dot(s.p1 - s.p2 , p - s.p2) < 0.0) return (p-s.p2).abs();
	return getdis_Li(s , p) ;
}

double getdis_ss(Segment s1 , Segment s2)  //线段与线段之间的距离
{
	if(intersect(s1.p1 , s1.p2 , s2.p1 , s2.p2)) return 0.0 ;
	return min(min(getdis_sp(s1 , s2.p1) , getdis_sp(s1 , s2.p2)) , min(getdis_sp(s2 , s1.p1) , getdis_sp(s2 , s1.p2))) ;
}

Point getCrossPoint(Segment s1 , Segment s2) //线段的交点，返回的是点
{
    Vector base = s2.p2 - s2.p1 ;
    double d1 = abs(cross(base , s1.p1 - s2.p1)) ;
    double d2 = abs(cross(base , s1.p2 - s2.p2)) ;
    double t = d1 / (d1 + d2) ;
    return s1.p1 + (s1.p2 - s1.p1) * t ;
}

bool intersect(Circle c , Line l)
{
    if(getdis_Li(l , c.c) > c.r)
    {
        return false ;
    }
    else
    {
        return true ;
    }
}

pair<Point, Point> getCrossPoints(Circle c, Line l)  //直线与圆的交点
{
    assert(intersect(c , l)) ;
    Vector pr = project(l , c.c) ;
    Vector e = (l.p2 - l.p1) / (l.p2 - l.p1).abs() ;
    double base = sqrt(c.r * c.r - (pr - c.c).norm()) ;
    return make_pair(pr + e * base, pr - e * base);
}

int main()
{
    Circle c ;
    cin >> c.c.x >> c.c.y >> c.r ;
    Line l ;
    int t ;
    cin >> t ;
    pair<Point , Point> p ;
    while(t --)
    {
        cin >> l.p1.x >> l.p1.y >> l.p2.x >> l.p2.y ;
        p = getCrossPoints(c , l) ;
        if(p.first.x < p.second.x || (p.first.x == p.second.x && p.first.y <= p.second.y))
        {
            printf("%.6lf %.6lf %.6lf %.6lf\n" , p.first.x , p.first.y , p.second.x , p.second.y) ;
        }
        else
        {
            printf("%.6lf %.6lf %.6lf %.6lf\n", p.second.x, p.second.y, p.first.x, p.first.y);
        }
    }
    return 0 ;
}