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>

using namespace std;
using uint = unsigned int;
using ll = long long;
using ull = unsigned long long;
constexpr ll TEN(int n) { return (n==0) ? 1 : 10*TEN(n-1); }
template<class T> using V = vector<T>;
template<class T> using VV = V<V<T>>;

template<class T> ostream& operator<<(ostream& os, const V<T> &v) {
    os << "["; for (auto p: v) os << p << ", "; os << "]";
    return os;
}

using D = double;
const D PI = acos(D(-1)), EPS = 1e-10;

int sgn(D a) { return (abs(a) <= EPS) ? 0 : (a < 0 ? -1 : 1); }
int sgn(D a, D b) { return sgn(a-b); }
//relative sign
// int rsgn(D a, D f) {
//     if (abs(a) <= f*EPS) return 0;
//     return (a < 0) ? -1 : 1;
// }
struct Pt2 {
    D x, y;
    Pt2() {}
    Pt2(D _x, D _y) : x(_x), y(_y) {}
    Pt2 operator+(const Pt2 &r) const { return Pt2(x+r.x, y+r.y); }
    Pt2 operator-(const Pt2 &r) const { return Pt2(x-r.x, y-r.y); }
    Pt2 operator*(const Pt2 &r) const { return Pt2(x*r.x-y*r.y, x*r.y+y*r.x); }

    Pt2 operator*(const D &r) const { return Pt2(x*r, y*r); }
    Pt2 operator/(const D &r) const { return Pt2(x/r, y/r); }

    Pt2& operator+=(const Pt2 &r) { return *this=*this+r; }
    Pt2& operator-=(const Pt2 &r) { return *this=*this-r; }
    Pt2& operator*=(const Pt2 &r) { return *this=*this*r; }
    Pt2& operator*=(const D &r) { return *this=*this*r; }
    Pt2& operator/=(const D &r) { return *this=*this/r; }
    
    Pt2 operator-() const { return Pt2(-x, -y); }

    bool operator<(const Pt2 &r) const { return 2*sgn(x, r.x)+sgn(y, r.y)<0; }
    bool operator==(const Pt2 &r) const { return sgn((*this-r).rabs()) == 0; }

    D norm() const { return x*x + y*y; }
    D abs() const { return sqrt(norm()); }
    D rabs() const { return max(std::abs(x), std::abs(y)); } // robust abs
    D arg() const { return atan2(y, x); }

    pair<D, D> to_pair() const { return make_pair(x, y); }
    static Pt2 polar(D le, D th) { return Pt2(le*cos(th), le*sin(th)); }
};
ostream& operator<<(ostream& os, const Pt2 &p) {
    return os << "P(" << p.x << ", " << p.y << ")";
}
using P = Pt2;

struct L {
    P s, t;
    L() {}
    L(P _s, P _t) : s(_s), t(_t) {}
    P vec() const { return t-s; }
    D abs() const { return vec().abs(); }
    D arg() const { return vec().arg(); }
};
ostream& operator<<(ostream& os, const L &l) {
    return os << "L(" << l.s << ", " << l.t << ")";
}

D cross(P a, P b) { return a.x*b.y - a.y*b.x; }
D dot(P a, P b) { return a.x*b.x + a.y*b.y; }
// cross(a, b) is too small?
int sgncrs(P a, P b) {
    D cr = cross(a, b);
    if (abs(cr) <= (a.rabs() + b.rabs()) * EPS) return 0;
    return (cr < 0) ? -1 : 1;
}

// -2, -1, 0, 1, 2 : front, clock, on, cclock, back
int ccw(P b, P c) {
    int s = sgncrs(b, c);
    if (s) return s;
    if (!sgn(c.rabs()) || !sgn((c-b).rabs())) return 0;
    if (dot(b, c) < 0) return 2;
    if (dot(-b, c-b) < 0) return -2;
    return 0;
}
int ccw(P a, P b, P c) { return ccw(b-a, c-a); }
int ccw(L l, P p) { return ccw(l.s, l.t, p); }

using Pol = V<P>;

D area2(const Pol &pol) {
    D u = 0;
    if (!pol.size()) return u;
    P a = pol.back();
    for (auto b: pol) u += cross(a, b), a = b;
    return u;
}

// 0: outside, 1: on line, 2: inside
int contains(const Pol &pol, P p) {
    if (!pol.size()) return 0;
    int in = -1;
    P _a, _b = pol.back();
    for (int i = 0; i < int(pol.size()); i++) {
        _a = _b; _b = pol[i];
        P a = _a, b = _b;
        if (ccw(a, b, p) == 0) return 1;
        if (a.y > b.y) swap(a, b);
        if (!(a.y <= p.y && p.y < b.y)) continue;
        if (sgn(a.y, p.y) ? (cross(a-p, b-p) > 0) : (a.x > p.x)) in *= -1;
    }
    return in + 1;
}

// p must be sorted and uniqued!
Pol convex_down(const V<P> &ps) {
    assert(ps.size() >= 2);
    Pol dw;
    for (P d: ps) {
        size_t n;
        while ((n = dw.size()) > 1) {
            if (ccw(dw[n-2], dw[n-1], d) != -1) break; // line上も取る
//            if (ccw(dw[n-2], dw[n-1], d) == 1) break;
            dw.pop_back();
        }
        dw.push_back(d);
    }
    return dw;
}

Pol convex(Pol ps) {
    sort(begin(ps), end(ps));
    ps.erase(unique(begin(ps), end(ps)), end(ps));
    if (ps.size() <= 1) return ps;
    Pol dw = convex_down(ps);
    reverse(begin(ps), end(ps));
    Pol up = convex_down(ps);
    dw.insert(begin(dw), begin(up) + 1, end(up) - 1);
    return dw;
}


int crossLL(const L &l, const L &m, P &r) {
    D cr1 = cross(l.vec(), m.vec()), cr2 = cross(l.vec(), l.t - m.s);
    if (sgncrs(l.vec(), m.vec()) == 0) {
        r = l.s;
        if (sgncrs(l.vec(), l.t - m.s)) return 0;
        return -1;
    }
    r = m.s + m.vec() * (cr2 / cr1);
    return 1;
}

// スパゲッティソースをパクった、直線の左側の多角形を返却する   
Pol convex_cut(const Pol &po, const L &l) {
    if (!po.size()) return Pol{};
    int n = int(po.size());
    Pol q;
    P a, b = po.back();
    for (int i = 0; i < n; i++) {
        a = b; b = po[i];
        if ((ccw(l, a)%2)*(ccw(l, b)%2) < 0) {
            P buf;
            crossLL(l, L(a, b), buf);
            q.push_back(buf);
        }
        if (ccw(l, b) != -1) q.push_back(b);
    }
    return q;
}


int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << setprecision(20) << fixed;
 
    int n;
    cin >> n;
    V<P> p(n);
    for (int i = 0; i < n; i++) {
        D x, y;
        cin >> x >> y;
        p[i] = P(x, y);
    }
    int q;
    cin >> q;
    for (int ph = 0; ph < q; ph++) {
        L l;
        D x, y;
        cin >> x >> y; l.s = P(x, y);
        cin >> x >> y; l.t = P(x, y);
        cout << area2(convex_cut(p, l)) / 2 << endl;
    }
 
    return 0;
}