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 <cstring>
#include <cmath>
#include <cassert>
#pragma warning(disable: 4996)
#define EPS (1e-10)
#define equals(a,b) (fabs(a)-fabs(b)<EPS)

using namespace std;

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(x * a, y * a); }
	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;

///Program 16.3 表示线段的结构体
struct Segment {
	Point p1, p2;
};

///Program 16.4 表示直线的结构体
typedef Segment Line;

class Circle {
public:
	Point c;
	double r;
	Circle(Point c = Point(), double r = 0.0) :c(c), r(r) {}
};


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

int ccw(Point p0, Point p1, Point p2) {
	Vector a = p1 - p0;
	Vector b = p2 - p0;
	if (cross(a, b) > EPS) {//printf("COUNTER_CLOCKWISE\n");
		return 1;
	}///p2位于p0-->p1的逆时针方向
	if (cross(a, b) < -EPS) {//printf("CLOCKWISE\n");
		return -1;
	}///p2位于p0-->p1的顺时针方向
	if (dot(a, b) < -EPS) {//printf("ONLINE_BACK\n");
		return 2;
	}///p2位于线段p0p1后方
	if (a.norm() < b.norm()) { //printf("ONLINE_FRONT\n");
		return -2;
	}///p2位于线段p0p1前方

//  printf("ON_SEGMENT\n");///p2位于线段p0p1的中间
	return 0;
}

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);
}

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

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

Point project(Segment s, Point p) {
	Vector base = s.p2 - s.p1;
	double r = dot(p - s.p1, base) / base.norm();
	//   return s.p1 + (base*r);
	Point ans;
	ans.x = s.p1.x + (base * r).x;
	ans.y = s.p1.y + (base * r).y;
	return ans;
}

pair<Point, Point> getCrossPoints(Circle c, Line l) {
	// assert(intersect(c.c,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);
}

int main()
{
	Circle c;
	scanf("%lf%lf%lf", &c.c.x, &c.c.y, &c.r);
	int q;
	scanf("%d", &q);
	for (int i = 0; i < q; i++) {
		Line l;
		scanf("%lf%lf%lf%lf", &l.p1.x, &l.p1.y, &l.p2.x, &l.p2.y);
		pair<Point, Point> p = getCrossPoints(c, l);
		if((p.first.x<p.second.x)||(p.first.x==p.second.x&&p.first.y<p.second.y))
			printf("%.8f %.8f %.8f %.8f\n", p.first.x, p.first.y, p.second.x, p.second.y);
		else 
			printf("%.8f %.8f %.8f %.8f\n", p.second.x, p.second.y, p.first.x, p.first.y);
		
	}
	return 0;
}