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<cstdio>
#include<vector>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;

#define mp make_pair
#define fi first
#define se second
#define pb push_back
typedef double db;
const db eps=1e-6;
const db pi=acos(-1.0);

int sign(db k){//-----------
    if (k>eps) return 1; else if (k<-eps) return -1; return 0;
}
int cmp(db k1,db k2){return sign(k1-k2);}  //-----------
struct point{
    db x,y;
    // point (double _x =0,double _y=0){x=_x;y=_y;}//------
    // point operator + (const point &k1) const {return point(k1.x+x,k1.y+y);}//------
    // point operator - (const point &k1) const {return point(x-k1.x,y-k1.y);}//------
    // point operator * (db k1) const  {return point(x*k1,y*k1);};//--------
    // point operator / (db k1) const{return point(x/k1,y/k1);}
    point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};}  //两个点相加-----
    point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};}  //两个点相减------
    point operator * (db k1) const{return (point){x*k1,y*k1};}  //与原点的距离扩大k1倍-------
    point operator / (db k1) const{return (point){x/k1,y/k1};}  //与原点的距离缩小k1倍
    int operator == (const point &k1) const{return cmp(x,k1.x)==0&&cmp(y,k1.y)==0;}  //判断两个点相等
    point turn(db k1){return (point){x*cos(k1)-y*sin(k1),x*sin(k1)+y*cos(k1)};} // 绕原点逆时针旋转
    point turn2(db k1, point p){return p+((*this)-p).turn(k1);}//绕p逆时针旋转k1度
    point turn90(){return (point){-y,x};}   //逆时针旋转90度
    //在点k1左边返回1 相等时在下边返回1  根据（x，y）排序-------------
    bool operator < (const point k1) const{
        int a=cmp(y,k1.y);
        if (a==-1) return 1; else if (a==1) return 0; else return cmp(x,k1.x)==-1;
    }
    db abs(){return sqrt(x*x+y*y);}//返回距离原点欧式距离
    db abs2(){return x*x+y*y;}  //返回距离原点欧式距离的平方
    db dis(point k1){return ((*this)-k1).abs();}    //返回两个点之间的欧式距离
    point unit(){db w=abs(); return (point){x/w,y/w};}  //返回单位向量
    void scan(){double k1,k2; scanf("%lf%lf",&k1,&k2); x=k1; y=k2;} //输入一个点--------
    void print(){printf("%.f %.f\n",x,y);}    //输出一个点--------
    db getw(){return atan2(y,x);}   //返回方位角  取值[-π，π]
    //如果 x>=0&&y>=0 不变，其他关于原点对称
    point getdel(){if (sign(x)==-1||(sign(x)==0&&sign(y)==-1)) return (*this)*(-1); else return (*this);}
};
db cross(point k1,point k2){return k1.x*k2.y-k1.y*k2.x;}    //叉积-----
struct line{
    // p[0]->p[1]  p[0]起点
    point p[2];
    line(point k1,point k2){p[0]=k1; p[1]=k2;}//---------
    point& operator [] (int k){return p[k];}//-------
    int include(point k){return sign(cross(p[1]-p[0],k-p[0]))>0;} //在右侧-----
    point dir(){return p[1]-p[0];} //--------
    line push(){ // 向内 ( 右手边 ) 平移 eps
        const db eps = 1e-6;
        point delta=(p[1]-p[0]).turn90().unit()*eps;
        return {p[0]-delta,p[1]-delta};
    }
};

int clockwise(point k1,point k2,point k3){// k1 k2 与 k1 k3 逆时针 1 顺时针 -1 否则 0  ---------
    return sign(cross(k2-k1,k3-k1));
}
point getLL(point k1,point k2,point k3,point k4){//直线 k1-k2 与 k3-k4 的交点  (调用前保证两直线有唯一交点-----
    db w1=cross(k1-k3,k4-k3),w2=cross(k4-k3,k2-k3); return (k1*w2+k2*w1)/(w1+w2);
}
vector<point> convexcut(vector<point>A,point k1,point k2){
    // 保留 k1,k2,p 逆时针的所有点
    int n=A.size(); A.push_back(A[0]); vector<point>ans;
    for (int i=0;i<n;i++){
        int w1=clockwise(k1,k2,A[i]),w2=clockwise(k1,k2,A[i+1]);
        if (w1>=0) ans.push_back(A[i]);
        if (w1*w2<0) ans.push_back(getLL(k1,k2,A[i],A[i+1]));
    }
    return ans;
}
db area(vector<point> A){ // 多边形的面积，多边形用 vector<point> 表示 , 逆时针
    db ans=0;
    for (int i=0;i<A.size();i++) ans+=cross(A[i],A[(i+1)%A.size()]);
    return ans/2;
}
vector < point > ans;
int main()
{
    int n;
    ans.clear();
    scanf("%d", &n);
    for(int i = 0; i < n;i++)
    {
        point a;
        a.scan();
        ans.push_back(a);
    }
    int k;
    scanf("%d", &k);
    while(k--)
    {
        point a,b;
        a.scan();
        b.scan();
        vector < point > ans_ = convexcut(ans, a,b);
        db areas = area(ans_);
        printf("%.12f\n",areas);
    }
    return 0;
}