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<bits/stdc++.h>
#include<iostream>
#include<stdio.h>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;
typedef double db;
const db eps = 1e-15, pi = acos(-1);
int sign(db x) {return x < -eps ? -1 : x > eps;}
int cmp(db x, db y) {return sign(x - y);}
int intersect(db l1, db r1, db l2, db r2) {
    if (l1 > r1) swap(l1, r1); if (l2 > r2) swap(l2, r2); return cmp(r1, l2) != -1 && cmp(r2, l1) != -1;
}
int inmid(db k1, db k2, db k3) {return sign(k1 - k3) * sign(k2 - k3) <= 0;}//k3 in [k1,k2]?1:0
struct Point {
    db x, y;
    Point operator + (const Point & a)const {return Point{a.x + x, a.y + y};}
    Point operator - (const Point & a)const {return Point{x - a.x, y - a.y};}
    Point operator * (db a) const {return Point{x * a, y * a};}
    Point operator / (db a) const {return Point{x / a, y / a};}
    bool operator < (const Point p) const {int a = cmp(x, p.x); if (a) return a == -1; return cmp(y, p.y) == -1;}
    bool operator == (const Point & a) const {return cmp(x, a.x) == 0 && cmp(y, a.y) == 0;}
    db abs() {return sqrt(x * x + y * y);}
    db dis(Point p) {return ((*this) - p).abs();}
    int getP() const {return sign(y) == 1 || (sign(y) == 0 && sign(x) == -1);}
    void input() {scanf("%lf%lf", &x, &y);}
};
db cross(Point p1, Point p2) {return p1.x * p2.y - p1.y * p2.x;}
db cross(Point p0, Point p1, Point p2) {return cross(p1 - p0, p2 - p0);}
db dot(Point p1, Point p2) {return p1.x * p2.x + p1.y * p2.y;}
int inmid(Point k1, Point k2, Point k3) {return inmid(k1.x, k2.x, k3.x) && inmid(k1.y, k2.y, k3.y);}
bool compareangle(Point p1, Point p2) {//Polar Angle Sort
    return p1.getP() < p2.getP() || (p1.getP() == p2.getP() && sign(cross(p1, p2)) > 0);
}
int clockwise(Point p1, Point p2, Point p3) { // p1 p2 p3 anticlockwise:1 clockwise:-1 others:0
    return sign(cross(p1, p2, p3));
}
struct Line {
    Point s, e;
    void input() {scanf("%lf%lf%lf%lf", &s.x, &s.y, &e.x, &e.y);}
    Point vec() {return e - s;}
    db length() {return sqrt(dot(s - e, s - e));}
    db length2() {return dot(s - e, s - e);}
};
int onS(Line l, Point p) {// On Seg?
    return inmid(l.s, l.e, p) && sign(cross(l.s - p, l.e - l.s)) == 0;
}
bool checkLL(Line l1, Line l2) {
    return cmp(cross(l1.s, l2.s, l2.e), cross(l1.e, l2.s, l2.e)) != 0;
}
bool checkLS(Line l1, Line l2) {//Intersection of Line l1 and Seg l2?
    return sign(cross(l2.s, l1.s, l1.e)) * sign(cross(l2.e, l1.s, l1.e)) <= 0;
}
int checkSS(Line l1, Line l2) {//Intersection of Two Seg?1:0
    return intersect(l1.s.x, l1.e.x, l2.s.x, l2.e.x) && intersect(l1.s.y, l1.e.y, l2.s.y, l2.e.y) && checkLS(l1, l2) && checkLS(l2, l1);
}
Point project(Line l, Point p) {
    return l.s + l.vec() * dot(p - l.s, l.vec()) / l.length2();
}
Point reflect(Line l, Point p) {//Mirror Point
    return project(l, p) * 2 - p;
}
Point getLL(Line l1, Line l2) {//Intersection Point of Line l1,l2
    db w1 = cross(l2.s, l1.s, l2.e), w2 = cross(l2.s, l2.e, l1.e); return (l1.s * w2 + l1.e * w1) / (w1 + w2);
}
db disSP(Line l, Point p) {
    Point p2 = project(l, p);
    if (inmid(l.s, l.e, p2)) return p.dis(p2); else return min(p.dis(l.s), p.dis(l.e));
}
db disSS(Line l1, Line l2) {
    if (checkSS(l1, l2)) return 0;
    return min(min(disSP(l1, l2.s), disSP(l1, l2.e)), min(disSP(l2, l1.s), disSP(l2, l1.e)));
}
db area(vector<Point> A) {//Anticlockwise
    db ans = 0;
    for (int i = 0; i < A.size(); i++) ans += cross(A[i], A[(i + 1) % A.size()]);
    return ans / 2;
}
int contain(vector<Point>A, Point p) {//2:in 1:on 0:out
    int ans = 0; A.push_back(A[0]);
    for (int i = 1; i < A.size(); i++) {
        Line l = {A[i - 1], A[i]};
        if (onS(l, p)) return 1; if (cmp(l.s.y, l.e.y) > 0) swap(l.s, l.e);
        if (cmp(l.s.y, p.y) >= 0 || cmp(l.e.y, p.y) < 0) continue;
        if (sign(cross(l.e, l.s, p)) < 0)ans ^= 1;
    }
    return ans << 1;
}
bool checkconvex(vector<Point>A) { //anticlock
    int n = A.size(); A.push_back(A[0]); A.push_back(A[1]);
    for (int i = 0; i < n; i++) if (sign(cross(A[i], A[i + 1], A[i + 2])) == -1) return 0;
    return 1;
}
vector<Point> ConvexHull(vector<Point>A, int flag = 1) { // flag=0 不严格 flag=1 严格
    int n = A.size(); vector<Point>ans(n * 2);
    sort(A.begin(), A.end()); int now = 0; if (n <= 1)return A;
    for (int i = 0; i < n; ans[now++] = A[i++])
        while (now > 1 && sign(cross(ans[now - 2], ans[now - 1], A[i])) < flag)--now;
    for (int i = n - 2, pre = now; i >= 0; ans[now++] = A[i--])
        while (now > pre && sign(cross(ans[now - 2], ans[now - 1], A[i])) < flag)--now;
    ans.resize(now - 1); return ans;
}
db convexDiameter(vector<Point>A) {
    int n = A.size(); if (n <= 1)return 0;
    int is = 0, js = 0; for (int k = 1; k <= n; ++k)is = A[k] < A[is] ? k : is, js = A[js] < A[k] ? k : js;
    int i = is, j = js; db ret = A[i].dis(A[j]); do {
        if (sign(cross(A[(i + 1) % n] - A[i], A[(j + 1) % n] - A[j])) >= 0)(++j) %= n;
        else (++i) %= n;
        ret = max(ret, A[i].dis(A[j]));
    } while (i != is || j != js);
    return ret;
}
vector<Point> convexcut(vector<Point>A, Line l) { // 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(l.s, l.e, A[i]), w2 = clockwise(l.s, l.e, A[i + 1]);
        if (w1 >= 0) ans.push_back(A[i]);
        if (w1 * w2 < 0) ans.push_back(getLL(l,Line{A[i], A[i + 1]}));
    }
    return ans;
}
int main() {
    int n; cin >> n;vector<Point> A(n);
    for (int i = 0; i < A.size(); ++i)A[i].input();
    int Q; cin >> Q; while (Q--) {
        Line l; l.input();
        vector<Point> B = convexcut(A, l);
        printf("%.10lf\n",area(B));
    }
}