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:
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <queue>
#include <cmath>
using namespace std;
#define EPS  (1e-10)
#define equals(a,b) (fabs((a) - (b)) < EPS)

// 点类
class Point {
public :
	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 a) { return Point(x * a, y * a); }
	Point operator / (double a) { return Point(x / a, y / a); }
	
	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;
	}
};
// 线段类
class Segment {
public:
	Point p1, p2;
	Segment() {};
	Segment(Point p1, Point p2) :p1(p1), p2(p2) {};
};
// 圆类
class Circle {
public:
	Point c;
	double r;
	Circle() {};
	Circle(Point c, double r) :c(c), r(r) {}
};
// 定义向量
typedef Point Vector;
// 定义直线
typedef Segment Line;
// 定义多边形
typedef vector<Point> Polygon;

/***************************点、向量****************************/

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

// 向量的内积
double dot(Point a, Point b) {
	return a.x * b.x + a.y * b.y;
}
// 向量的外积
double cross(Point a, Point b) {
	return a.x * b.y - a.y * b.x;
}

// 向量a，b是否正交 <==> 内积为0
bool isOrthogonal(Vector a, Vector b) {
	return equals(dot(a, b), 0.0);
}
bool isOrthogonal(Point a1, Point a2, Point b1, Point b2) {
	return equals(dot(a1 - a2, b1 - b2), 0.0);
}

// 向量a，b是否平行 <==> 外积为0
bool isParallel(Vector a, Vector b) {
	return equals(cross(a, b), 0.0);
}
bool isParallel(Point a1, Point a2, Point b1, Point b2) {
	return equals(cross(a1 - a2, b1 - b2), 0.0);
}

// 点p在线段s上的投影
Point project(Segment s, Point p) {
	Vector base = s.p2 - s.p1;
	double r = dot(p - s.p1, base) / norm(base);
	return s.p1 + base * r ;
}

//以线段s为对称轴与点p成线对称的点
Point reflect(Segment s, Point p) {
	return p + (project(s, p) - p) * 2.0;
}

// 点a到点b的距离
double getDistance(Point a, Point b) {
	return abs(a - b);
}

// 线段l和点p的距离
double getDistanceLP(Line l, Point p) {
	return abs( cross(l.p2 - l.p1, p - l.p1) / abs(l.p2 - l.p1) );
}

// 线段s与点p的距离
double getDistanceSP(Segment s, Point p) {
	if (dot(s.p2 - s.p1, p - s.p1) < 0.0)
		return abs(p - s.p1);
	if (dot(s.p1 - s.p2, p - s.p2) < 0.0)
		return abs(p - s.p2);
	return getDistanceLP(s, p);
}



/*************************线段********************************/
// 线段s1，s2是否正交 <==> 内积为0
bool isOrthogonal(Segment s1, Segment s2) {
	return equals(dot(s1.p2 - s1.p1, s2.p2 - s2.p1), 0.0);
}

// 线段s1，s2是否平行 <==> 外积为0
bool isParallel(Segment s1, Segment s2) {
	return equals(cross(s1.p2 - s1.p1, s2.p2 - s2.p1), 0.0);
}

// 逆时针方向ccw（Counter-Clockwise）
static const int COUNTER_CLOCKWISE = 1;
static const int CLOCKWISE = -1;
static const int ONLINE_BACK = 2;
static const int ONLINE_FRONT = -2;
static const int ON_SEGMENT = 0;

int ccw(Point p0, Point p1, Point p2) {
	Vector a = p1 - p0;
	Vector b = p2 - p0;
	if (cross(a, b) > EPS) return COUNTER_CLOCKWISE;
	if (cross(a, b) < -EPS) return CLOCKWISE;
	if (dot(a, b) < -EPS) return ONLINE_BACK;
	if (norm(a) < norm(b)) return ONLINE_FRONT;
	return ON_SEGMENT;
}

// 判断线段p1p2和线段p3p4是否相交
bool intersect(Point p1, Point p2, Point p3, Point p4) {
	return (ccw(p1, p2, p3) * ccw(p1, p2, p4) <= 0 &&
		ccw(p3, p4, p1) * ccw(p3, p4, p2) <= 0);
}

//判断线段s1和s2是否相交
bool intersect(Segment s1, Segment s2) {
	return intersect(s1.p1, s1.p2, s2.p1, s2.p2);
}

// 线段s1和线段s2的距离
double getDistance(Segment s1, Segment s2) {
	// 相交
	if (intersect(s1, s2))
		return 0.0;
	return min(min(getDistanceSP(s1, s2.p1), getDistanceSP(s1, s2.p2)),
		min(getDistanceSP(s2, s1.p1), getDistanceSP(s2, s1.p2)));
}

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

/***************************圆****************************/

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

// 圆c1和圆c2的交点
double arg(Vector p) { return atan2(p.y, p.x); }
Vector polar(double a, double r) { return Point(cos(r) * a, sin(r) * a); }

pair<Point, Point> getCrossPoints(Circle c1, Circle c2) {
	double d = abs(c1.c - c2.c);
	double a = acos((c1.r * c1.r + d * d - c2.r * c2.r) / (2 * c1.r * d));
	double t = arg(c2.c - c1.c);
	return make_pair(c1.c + polar(c1.r, t + a), c1.c + polar(c1.r, t - a));
}

/***************************多边形****************************/
// 点的内包
/*
	IN 2
	ON 1
	OUT 0
*/
int contains(Polygon g, Point p) {
	int n = g.size();
	bool x = false;
	for (int i = 0; i < n; i++) {
		Point a = g[i] - p, b = g[(i + 1) % n] - p;
		if (abs(cross(a, b)) < EPS && dot(a, b) < EPS) return 1;
		if (a.y > b.y) swap(a, b);
		if (a.y < EPS && EPS < b.y && cross(a, b) > EPS)
			x = !x;
	}
	return (x ? 2 : 0);
}

// 凸包
Polygon andrewScan(Polygon s) {
	Polygon u, l;
	int len = s.size();
	if (len < 3) return s;
	// 以x，y为基准升序排序
	sort(s.begin(), s.end());
	// 将x值最小的两个点添加到u
	u.push_back(s[0]);
	u.push_back(s[1]);

	// 将x值最大的两个点添加到l
	l.push_back(s[len - 1]);
	l.push_back(s[len - 2]);

	// 构建凸包上部
	for (int i = 2; i < len; i++) {
		for (int j = u.size(); j >= 2 && ccw(u[j - 2], u[j - 1], s[i]) != CLOCKWISE; j--)
			u.pop_back();
		u.push_back(s[i]);
	}

	// 构建凸包下部
	for (int i = len - 3; i >= 0; i--) {
		for (int j = l.size(); j >= 2 && ccw(l[j - 2], l[j - 1], s[i]) != CLOCKWISE; j--)
			l.pop_back();
		l.push_back(s[i]);
	}

	// 按顺时针方向生成凸包的点的序列
	reverse(l.begin(), l.end());
	for (int i = u.size() - 2; i >= 1; i--)
		l.push_back(u[i]);

	return l;
}


int main() {
	Circle c;
	Line l;
	int n;
	scanf("%lf %lf %lf", &c.c.x, &c.c.y, &c.r);
	scanf("%d", &n);
	while (n--) {
		scanf("%lf %lf %lf %lf", &l.p1.x, &l.p1.y,
			&l.p2.x, &l.p2.y);
		pair<Point, Point> ans = getCrossPoints(c, l);
		if (ans.first < ans.second)
			printf("%.10lf %.10lf %.10lf %.10lf\n", ans.first.x, ans.first.y,
				ans.second.x, ans.second.y);
		else
			printf("%.10lf %.10lf %.10lf %.10lf\n", ans.second.x, ans.second.y,
				ans.first.x, ans.first.y);
	}
	return 0;
}