Problem: <image>


As shown in the figure above, cut a convex polygon g by a line p1p2 and print the area of the cut polygon which is on the left-hand side of the line.

g is represented by a sequence of points p1, p2,..., pn where line segments connecting pi and pi+1 (1 ≤ i ≤ n−1) are sides of the convex polygon. The line segment connecting pn and p1 is also a side of the polygon.

Constraints

* 3 ≤ n ≤ 100
* 1 ≤ q ≤ 100
* -10000 ≤ xi, yi ≤ 10000
* -10000 ≤ p1x,p1y,p2x,p2y ≤ 10000
* No point in g will occur more than once.
* p1 ≠ p2

Input

The input is given in the following format:


g (the sequence of the points of the polygon)
q (the number of queries = the number of target lines)
1st query
2nd query
:
qth query


g is given as a sequence of points p1,..., pn in the following format:


n
x1 y1
x2 y2
:
xn yn


The first integer n is the number of points. The coordinate of the i-th point pi is given by two integers xi and yi. The coordinates of points are given in the order of counter-clockwise visit of them. Note that all interior angles of given convex polygons are less than or equal to 180.

For each query, a line represented by two points p1 and p2 is given. The coordinates of the points are given by four integers p1x, p1y, p2x and p2y.

Output

For each query, print the area of the cut polygon. The output values should be in a decimal fraction with an error less than 0.00001.

Example

Input

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


Output

2.00000000
4.00000000

Solution (in CPP):

Solution:
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include<assert.h>
#include<vector>

using namespace std;

#define EPS (1e-10)
#define equals(a,b) (fabs((a)-(b))<EPS)

inline int sign(const double &x){
    if(x>EPS) return 1;
    else if(x<-EPS) return -1;
    return 0;
}

/*   点  */
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 k) { return Point(k*x,k*y); }
    Point operator / (double k) { return Point(x/k,y/k); }

    //向量的大小
    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;
    }
};
/*   向量  */
typedef Point Vector;

//向量a和b的内积
double dot(Vector a,Vector b){
    return a.x*b.x+a.y*b.y;
}
//向量a和b的外积
double cross(Vector a,Vector b){
    return a.x*b.y-a.y*b.x;
}
struct Segment{
    Point p1,p2;
};

/*   直线  */
typedef Segment Line;

//判断p2与向量p1-p0的位置关系

const int COUNTER_CLOCKWISE = -1;   //逆时针
const int CLOCKWISE = 1;            //顺时针
const int ON_SEGMENT = 0;           //在线段上
const int ONLINE_BACK = 2;          //p2在向量p1-p0的反方向延长线上
const int ONLINE_FRONT = -2;        //p2在向量p1-p0的正方向延长线上

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(a.norm()<b.norm()) return ONLINE_FRONT;
    return ON_SEGMENT;
}

/*   多边形  */
typedef vector<Point> Polygon;
/* *******凸包*****(包含点集合p中所有点的最小凸多边形)
 * 输出凸多边形最下端最左侧的顶点为起点,按逆时针方向依次输出坐标。
 * 安德鲁算法 O(n*logn)
 */

double Area(Point p0,Point p1,Point p2){
    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}
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;
}
Polygon andrewscanf(Polygon s){
    int n=s.size();
    sort(s.begin(),s.end());    //x从小到大，x相同的y从小到大
    Polygon u;
    //构建凸包上部
    for(int i=0;i<n;i++){
        for(int j=u.size();j>=2&&Area(u[j-2],u[j-1],s[i]) <= EPS;j--){
            u.pop_back();
        }
        u.push_back(s[i]);
    }
    //构建凸包下部
    int tmp=u.size()+1;
    for(int i=n-2;i>=0;i--){
        for(int j=u.size();j>=tmp&&Area(u[j-2],u[j-1],s[i]) <= EPS;j--){
            u.pop_back();
        }
        u.push_back(s[i]);
    }
    //按顺序生成凸包点的序列
    if(n>1) u.pop_back();               //开始点多加入了一次
    return u;
}

//旋转卡壳*-------》计算凸包直径
double RC(Polygon v){
    double res=0;
    int n=v.size();
    int k=1;
    for(int i=0;i<n;i++){
        while(sign(Area(v[i],v[(i+1)%n],v[(k+1)%n])-Area(v[i],v[(i+1)%n],v[k%n]))==1) k=(k+1)%n;
//        printf("*** %d\n",k);
        res=max(res,max((v[i]-v[k]).abs(),(v[(i+1)%n]-v[k]).abs()));
    }
    return res;
}

int main(){
    int n;
    scanf("%d",&n);
    Polygon g;
    for(int i=0;i<n;i++){
        double x,y;
        scanf("%lf%lf",&x,&y);
        g.push_back(Point(x,y));
    }
    int q;
    scanf("%d",&q);
    while(q--){
        Point a,b;
        scanf("%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y);
        Polygon tmp;
        for(int i=0;i<n;i++){
            Point t1=g[i],t2=g[(i+1)%n];
//            printf("*** %d %d\n",ccw(a,b,t1),ccw(a,b,t2));
            if(ccw(a,b,t1)!=1) tmp.push_back(t1);
            if(ccw(a,b,t1)*ccw(a,b,t2)==-1){
                tmp.push_back(getCrossPoint(Segment{t1,t2},Segment{a,b}));
                Point d=getCrossPoint(Segment{t1,t2},Segment{a,b});
            }
        }
        int len=tmp.size();
        double ans=0;
        for(int i=0;i<len;i++){
//            printf("---%.0f %.0f\n",tmp[i].x,tmp[i].y);
            Point t1=tmp[i],t2=tmp[(i+1)%len];
            ans+=cross(t1,t2);
        }
        ans/=2.0;
        printf("%.10f\n",ans);
    }
    return 0;
}