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;
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define rep(i, a, b) for(int i = (a); i < (b); ++i)
#define per(i, a, b) for(int i = (b) - 1; i >= (a); --i)
#define sz(a) (int)a.size()
#define de(c) cout << #c << " = " << c << endl
#define dd(c) cout << #c << " = " << c << " "
#define all(a) a.begin(), a.end()
#define pw(x) (1ll<<(x))
#define endl "\n"
typedef long long ll;
typedef double db;
typedef pair<int, int> pii;
typedef vector<int> vi;

typedef db T;
const db eps = 1e-9 , pi = acosl(-1.);
int sgn(T x){return (x>eps)-(x<-eps);}
struct P{
  T x,y; P(){} P(T x,T y):x(x),y(y){}
  P operator - (const P&b) const {return P(x-b.x,y-b.y);}
  P operator + (const P&b) const {return P(x+b.x,y+b.y);}
  T operator * (const P&b) const {return x*b.x+y*b.y;}
  T operator / (const P&b) const {return x*b.y-y*b.x;}
  P operator * (const T&k) const {return P(x*k,y*k);}
  P operator / (const T&k) const {return P(x/k,y/k);}
	bool operator < (const P&b) const {return sgn(x-b.x)?x<b.x:y<b.y;}
};
T norm(P a){return a*a;}
T abs(P a) {return sqrtl(norm(a));}
P proj(P p,P a,P b){return (b-a)*((p-a)*(b-a)/norm(b-a))+a;}
P reflect(P p,P a,P b){return proj(p,a,b)*2-p;}
T cross(P o,P a,P b){return (a-o)/(b-o);}
int crossOp(P o,P a,P b){return sgn(cross(o,a,b));}
bool onPS(P p,P s,P t){return sgn((t-s)/(p-s))==0&&sgn((p-s)*(p-t))<=0;}
struct L{ P s,t;L(){} L(P s,P t):s(s),t(t){}};
P insLL(L a,L b){ // line x line
	P s = a.s - b.s , v = a.t - a.s , w = b.t - b.s;
	db k1 = s / w , k2 = w / v;
	if(sgn(k2) == 0) return abs(b.s - a.s) < abs(b.t - a.s) ? b.s : b.t;
	return a.s + v * (k1 / k2);
}
bool isSS(L a,L b){ // seg x seg , replace x->y to accelerate
	T c1=(a.t-a.s)/(b.s-a.s),c2=(a.t-a.s)/(b.t-a.s);
	T c3=(b.t-b.s)/(a.s-b.s),c4=(b.t-b.s)/(a.t-b.s);
	return sgn(c1) * sgn(c2) <= 0 && sgn(c3) * sgn(c4) <= 0 &&
		sgn(max(a.s.x,a.t.x) - min(b.s.x,b.t.x)) >= 0 &&
		sgn(max(b.s.x,b.t.x) - min(a.s.x,a.t.x)) >= 0 &&
		sgn(max(a.s.y,a.t.y) - min(b.s.y,b.t.y)) >= 0 &&
		sgn(max(b.s.y,b.t.y) - min(a.s.y,a.t.y)) >= 0;
}
db disPL(P p,L a){return fabs((a.t-a.s)/(p-a.s)) / abs(a.t-a.s);}
db disPS(P p,L a){ // p x seg dis
	if(sgn((a.t-a.s)*(p-a.s)) == -1) return abs(p-a.s);
	if(sgn((a.s-a.t)*(p-a.t)) == -1) return abs(p-a.t);
	return disPL(p,a);
}
db disSS(L a,L b){ // seg x seg dis
	if(isSS(a,b)) return 0;
	return min(min(disPS(a.s,b),disPS(a.t,b)),min(disPS(b.s,a),disPS(b.t,a)));
}
typedef vector<P> polygon;
polygon convex(polygon A){ // counter-clockwise , < : <=180 , <= : <180
  int n=sz(A),m=0;
  polygon B;B.resize(n<<1);
  sort(all(A));
  rep(i,0,n){
    while(m > 1 && sgn((B[m-1]-B[m-2])/(A[i]-B[m-2]))<0) --m;
    B[m++]=A[i];
  }
  int k = m;
  per(i,0,n-1){
    while(m > k && sgn((B[m-1]-B[m-2])/(A[i]-B[m-2]))<0) --m;
    B[m++]=A[i];
  }
  B.resize(m);
  if(sz(B) > 1) B.pop_back();
  return B;
}
T area(polygon A) { // multiple 2 with integer type
	T res=0;
	rep(i,0,sz(A)) res+=A[i]/(A[(i+1)%sz(A)]);
	return fabs(res) / 2;
}
bool isconvex(polygon A){ // counter-clockwise
	bool ok=1;int n=sz(A);
	rep(i,0,2) A.pb(A[i]);
	rep(i,0,n) ok&=((A[i+1]-A[i])/(A[i+2]-A[i]))>=0;
	return ok;
}
int inPpolygon(P p,polygon A){ // -1 : on , 0 : out , 1 : in
	int res=0;
	rep(i,0,sz(A)){
		P u=A[i],v=A[(i+1)%sz(A)];
		if(onPS(p,u,v)) return -1;
		T cross = sgn((v-u)/(p-u)) , d1 = sgn(u.y-p.y) , d2 = sgn(v.y-p.y);
		if(cross > 0 && d1 <= 0 && d2 > 0) ++res;
		if(cross < 0 && d2 <= 0 && d1 > 0) --res;
	}
	return res != 0;
}
T diameter(polygon A) { // longest distance
  int n=sz(A);if(n <= 1) return 0;
  int l=0,r=0;rep(i,1,n) (A[i]<A[l])&&(l=i),(A[r]<A[i])&&(r=i);
  db res=abs(A[l]-A[r]);int i=l,j=r;
  do (++((A[(i+1)%n]-A[i])/(A[(j+1)%n]-A[j])>=0?j:i))%=n,
    res=max(res,abs(A[i]-A[j]));
  while(i!=l||j!=r);
  return res;
}
polygon convexCut(polygon A,P s,P t){ // counter-clockwise , left hand of st
	int n=sz(A);
	polygon B;
	rep(i,0,n){
		P u=A[i],v=A[(i+1)%n];
		int d1 = sgn((t-s)/(u-s)) , d2 = sgn((t-s)/(v-s));
		if(d1 >= 0) B.pb(u);
		if(d1 * d2 < 0) B.pb(insLL(L(u,v),L(s,t)));
	}
	return B;
}

P a, b, c, d;
polygon poly;

int main() {
	std::ios::sync_with_stdio(0);
	std::cin.tie(0);
	//	freopen("a.in", "r", stdin);
	//	freopen("a.out", "w", stdout);
	int q;
	cin >> q;
	cout << setiosflags(ios::fixed);
	cout << setprecision(10);
	while(q--) {
		cin >> a.x >> a.y;
		poly.pb(a);
	}
	cin >> q;
	while(q--) {
		cin >> a.x >> a.y >> b.x >> b.y;
		polygon A = convexCut(poly, a, b);
		cout << area(A) << endl;
	}
	return 0;
}