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 <vector>
#include <string>
#include <cmath>
#include <set>
#include <algorithm>
#include <array>
#include <complex>
#include <string>
#include <utility>
#include <map>
#include <queue>
#include <list>
#include <functional>
#include <numeric>
#include <stack>
#include <tuple>

using namespace std;
int dx[4] = { -1,0,1,0 };
int dy[4] = { 0,1,0,-1 };
const int INF = 100000000;
const long long LINF = 1000000000000000000;
const int MOD = (int)1e9 + 7;
const double EPS = 1e-6;
using pii = std::pair<int, int>;
using ll = long long;
using pLL = std::pair<ll, ll>;
#define SORT(v) std::sort(v.begin(), v.end())
#define RSORT(v) std::sort(v.rbegin(), v.rend())
using Point = complex<double>; //点
using Poly = vector<Point>;
#define X real() //実部
#define Y imag() //虚部
namespace std {
    bool operator<(const Point& a, const Point& b) {
        return a.X != b.X ? a.X < b.X : a.Y < b.Y;
    }

    bool cmp_y(const Point& a, const Point& b) {
        return a.Y != b.Y ? a.Y < b.Y : a.X < b.X;
    }
}
double dot(Point a, Point b) { //内積
    return (a.X * b.X + a.Y * b.Y);
}
using Line = pair<Point, Point>; //直線
Point Projection(Line s, Point p) { //点から直線への射影
    Point vase = s.second - s.first;
    p -= s.first;
    return s.first + dot(p, vase) / norm(vase) * vase;
}
Point Reflection(Line s, Point p) { //直線に対して線対称な点
    return p + (Projection(s, p) - p) * 2.0;
}
double cross(Point a, Point b) { //外積
    return a.X * b.Y - a.Y * b.X;
}
int ccw(Point a, Point b, Point c) { //直線に対する点の位置
    if (cross(b - a, c - a) > EPS)return 1;//左
    if (cross(b - a, c - a) < -EPS)return -1;//右
    if (dot(b - a, c - a) < -EPS)return 2;//線分の左端点の延長上
    if (abs(b - a) + EPS < abs(c - a))return -2;//線分の右端点の延長上
    return 0;//線分上
}
bool isOrthogonal(Line a, Line b) { //直行判定
    return dot(a.second - a.first, b.second - b.first) ? 0 : 1;
}
bool isParallel(Line a, Line b) { //平行判定
    return cross(a.second - a.first, b.second - b.first) ? 0 : 1;
}
bool isCross(Line a, Line b) { //交差判定
    if (ccw(a.first, a.second, b.first) * ccw(a.first, a.second, b.second) <= 0 and
        ccw(b.first, b.second, a.first) * ccw(b.first, b.second, a.second) <= 0) {
        return true;
    }
    return false;
}
Point CrossPoint(Line a, Line b) { //直線の交点
    double x, d;
    Point p1, p2, p3, p4;
    p1 = a.first;
    p2 = a.second;
    p3 = b.first;
    p4 = b.second;
    x = (p4.Y - p3.Y) * (p4.X - p1.X) - (p4.X - p3.X) * (p4.Y - p1.Y);
    d = (p4.Y - p3.Y) * (p2.X - p1.X) - (p4.X - p3.X) * (p2.Y - p1.Y);
    return p1 + x / d * (p2 - p1);
}
double PointLineDis(Line l, Point p) {
    Point a = l.first, b = l.second, c = p;
    return abs(cross(c - a, b - a)) / abs(b - a);
}
double PointSegDis(Line a, Point p) {
    Point p1 = a.first, p2 = a.second;
    if (dot(p2 - p1, p - p1) < EPS)return abs(p - p1);
    if (dot(p1 - p2, p - p2) < EPS)return abs(p - p2);
    return PointLineDis(a, p);
}
double SegSegDis(Line a, Line b) {
    double res = PointSegDis(a, b.first);
    cout << res << endl;
    res = min(res, PointSegDis(a, b.second));
    cout << res << endl;
    res = min(res, PointSegDis(b, a.first));
    cout << res << endl;
    res = min(res, PointSegDis(b, a.second));
    cout << res << endl;
    return res;
}
double PolyArea(Poly p) {
    if (p.size() < 3)return 0;
    double res = cross(p[p.size() - 1], p[0]);
    for (size_t i = 0; i < p.size() - 1; ++i) {
        res += cross(p[i], p[i + 1]);
    }
    return res / 2.0;
}
bool isConvex(Poly p) {
    for (size_t i = 0; i < p.size(); ++i) {
        if (ccw(p[i], p[(i + 1) % p.size()], p[(i + 2) % p.size()]) == -1)return false;
    }
    return true;
}
int inPoly(Poly& p, Point a) {
    int count = 0;
    Point cp;
    int n = (int)p.size();
    p.push_back(p[0]);
    Point b = Point(a.X + 1, a.Y);
    for (int i = 0; i < n; ++i) {
        Point s = p[i] - a, t = p[i + 1] - a;
        if (ccw(s, t, Point(0, 0)) == 0)return 1;
        if (s.Y > t.Y + EPS)swap(s, t);
        if ((s.Y * t.Y < 0 || (s.Y * t.Y<EPS && t.Y>EPS)) && cross(s, t) < EPS)count++;
    }
    if (count % 2)return 2;
    else return 0;
}
Poly ConvexHull(Poly v) {//凸包 by TAB
    int n = v.size(), k = 0;
    sort(v.begin(), v.end(), cmp_y);
    Poly r(2 * n);
    for (int i = 0; i < n; i++) {
        while (k > 1 && cross(r[k - 1] - r[k - 2], v[i] - r[k - 2]) < -EPS)k--;
        r[k++] = v[i];
    }
    for (int i = n - 2, t = k; i >= 0; i--) {
        while (k > t && cross(r[k - 1] - r[k - 2], v[i] - r[k - 2]) < -EPS)k--;
        r[k++] = v[i];
    }
    r.resize(k - 1);
    return r;
}
/*
Poly ConvexHull(Poly p) {
    int n = p.size(), k = 0;
    sort(p.begin(),p.end(),cmp_y);
    Poly r(2 * n);
    for (int i = 0; i < n; ++i) {
        while (k >= 2 && cross(r[k - 1]- r[k - 2], p[i]-r[k - 2]) <= 0)k--;
        r[k++] = p[i];
    }
    for (int i = n - 2, t = k + 1; i > 0; --i) {
        while (k >= t && cross(r[k - 1] - r[k - 2], p[i-1] - r[k - 2]) <= 0)k--;
        r[k++] = p[i - 1];
    }
    r.resize(k-1);
    return r;
}*/
double Caliper(const Poly &p) {//多角形の直径
    int n = p.size();
    if (n <= 1)return 0;
    if (n == 2)return abs(p[0] - p[1]);

    int i = 0, j = 0;
    for (int k = 0; k < n; ++k) {
        if (!(p[i] < p[k]))i = k;
        if (p[j] < p[k])j = k;
    }

    double res = 0.0;
    int si = i, sj = j;
    while (i != sj || j != si) {
        res = max(res, abs(p[i] - p[j]));
        if (cross(p[(i + 1) % n] - p[i], p[(j + 1) % n] - p[j]) < 0)i = (i + 1) % n;
        else j = (j + 1) % n;
    }
    return res;
}
// 全頂点に対してccwを見る
//左側にあれば解に入れる
Poly ConvexCut(Poly p, Line a) {
    int n = p.size();
    Poly res;
    for (int i = 0; i < n; ++i) {
        if (ccw(a.first, a.second, p[i]) != -1)res.push_back(p[i]);
        if (ccw(a.first, a.second, p[i])*ccw(a.first, a.second, p[(i + 1) % n]) ==-1) {
            res.push_back(CrossPoint(Line(p[i], p[(i + 1) % n]), a));
        }
    }
    return res;
}
int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    
    
    int n;
    cin >> n;
    Poly poly;
    for (int i = 0; i < n; ++i) {
        double x1, y1;
        cin >> x1 >> y1;
        Point p = Point(x1, y1);
        poly.push_back(p);
        //cout << p.X << p.Y << endl;
    }
    int q;
    cin >> q;
    for (int i = 0; i < q; ++i) {
        double x1, x2, y1, y2;
        cin >> x1 >> y1 >> x2 >> y2;
        Point p1 = Point(x1, y1);
        Point p2 = Point(x2, y2);
        Line a = Line(p1, p2);
        Poly ans = ConvexCut(poly, a);
        printf("%.9f\n", PolyArea(ans));
    }
    
    return 0;
}