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
#include<bits/stdc++.h>

using namespace std;
struct Point{
    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 norm(){return x*x+y*y;}//模的平方
    double ABS() {return sqrt(norm());}//模
};
//线段
struct Segment{
    Point p1,p2;
};
//点积
double dot(Point a,Point b){
    return a.x*b.x+a.y*b.y;
}
//叉积
double cross(Point a,Point b){
    return a.x*b.y-a.y*b.x;
}
Point Projection(Point p,Segment s){
    Point alp=p-s.p1;
    Point beta=s.p2-s.p1;
    double res=dot(alp,beta)/beta.norm();
    return s.p1+(beta*res);
}
//点到直线的距离
double PLDis(Point a,Segment s){
    double A=s.p1.y-s.p2.y;
    double B=s.p2.x-s.p1.x;
    double C=(s.p1.x-s.p2.x)*s.p1.y-(s.p1.y-s.p2.y)*s.p1.x;
    return fabs(A*a.x+B*a.y+C)/sqrt((A*A+B*B));
}
bool cmp(Point a,Point b){
    if(a.x==b.x) return a.y<b.y;
    return a.x<b.x;
}
//计算圆和直线的交点
vector<Point> CCL(Segment s,Point o,double r){
    vector<Point> res;
    Point x=Projection(o,s);
    double dis=PLDis(o,s);
    if(dis>r){//距离>r没有交点
        return res;
    }
    if(dis==r){
        res.push_back(x);
        res.push_back(x);
    }
    double beta=sqrt(r*r-dis*dis);
    Point pp=s.p2-s.p1;
    pp=pp/pp.ABS();//单位向量
    Point ans1=x-pp*beta;
    Point ans2=x+pp*beta;
    res.push_back(ans1);
    res.push_back(ans2);
    return res;
}
int main(){
    Point c;
    double r;
    scanf("%lf %lf %lf",&c.x,&c.y,&r);
    int q;
    scanf("%d",&q);
    while(q--){
        Segment s;
        scanf("%lf %lf %lf %lf",&s.p1.x,&s.p1.y,&s.p2.x,&s.p2.y);
        vector<Point> ans=CCL(s,c,r);
        sort(ans.begin(),ans.end(),cmp);
        printf("%.10lf %.10lf %.10lf %.10lf\n",ans[0].x,ans[0].y,ans[1].x,ans[1].y);
    }
    return 0;
}
