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>

typedef long double     lf;

using namespace std;

const lf EPS = 1e-9;
const lf oo = 1e15;

struct pt {
  lf x, y;
  pt( ) { }
  pt( lf x, lf y ) : x( x ), y ( y ) { }
};

inline lf x( pt P ) { return P.x; }
inline lf y( pt P ) { return P.y; }

istream& operator >> ( istream& in, pt& p ) {
  lf x,y; in >> x >> y;
  p = pt(x,y); return in;
}

ostream& operator << ( ostream& out, const pt& p ) {
  out << int(p.x) << " " << int(p.y);
  return out;
}

pt operator + ( const pt& A, const pt& B ) {
  return pt( x(A)+x(B), y(A)+y(B) );
}

pt operator - ( const pt& A, const pt& B ) {
  return pt( x(A)-x(B), y(A)-y(B) );
}


pt operator * ( const lf& B, const pt& A ) {
  return pt( x(A)*B, y(A)*B );
}

pt operator * ( const pt& A, const lf& B ) {
  return pt( x(A)*B, y(A)*B  );
}

inline lf dot( pt A, pt B ) {
  return x(A)*x(B) + y(A)*y(B);
}

inline lf norm( pt A ) {
  return x(A)*x(A) + y(A)*y(A);
}

inline lf abs( pt A ) {
  return sqrt( norm(A) );
}

inline lf dist ( pt A, pt B ) { return abs( B - A ); }

lf distToLine (pt p, pt A, pt B, pt &c) {
  lf u = dot( p-A , B-A ) / norm( B-A );
  c = A + u*( B-A );
  return dist( p , c );
}

pt refPoint(pt X, pt A, pt B) {
  pt aux; distToLine(X, A, B, aux);
  return X + lf(2.0)*(aux-X);
}

inline bool same ( lf a, lf b ) {
  return a+EPS > b && b+EPS > a;
}

inline lf cross( pt A, pt B ) {
  return x(A)*y(B) - y(A)*x(B);
}

///CHANGE
// 0  for collineal points ( angle = 0 )
// 1  for angle BAX counter clockwise
// -1 for angle BAX clockwise
inline int ccw (pt X, pt A, pt B) {
  lf c = cross( B-A, X-A );
  if( same( c, 0.0 ) ) { return 0; }
  if( c > EPS ) { return 1; }
  return -1;
}

///CHANGE
inline bool segContains ( pt X, pt A, pt B) {
  if ( !same ( 0, cross ( A-X, B-X ) ) ) return 0;
  return ( dot ( A-X, B-X ) < EPS );
}

inline bool parallel( pt A, pt B, pt C, pt D ) {
  return same ( 0, cross( B-A, D-C ) );
}

///NEW
inline bool ortho( pt A, pt B, pt C, pt D ) {
  return same ( 0, dot( B-A, D-C ) );
}

inline bool samePt ( pt A, pt B ) {
  return same ( x(A), x(B) ) && same ( y(A), y(B) );
}

pt linesIntersection ( pt A, pt B, pt C, pt D ) {
  lf x = cross ( C, D-C ) - cross ( A, D-C );
  x /= cross ( B-A, D-C );
  return A + x*(B-A);
}

inline bool collinearSegsIntersects ( pt A, pt B, pt C, pt D ) {
  return segContains(A,C,D) || segContains(B,C,D)
      || segContains(C,A,B) || segContains(D,A,B);
}

bool segmentsIntersect(pt A, pt B, pt C, pt D) {
  if( samePt(A,B) )
    return segContains( A, C, D );
  if( samePt(C,D) )
    return segContains( C, A, B );
  if( parallel(A,B,C,D) )
    return collinearSegsIntersects( A,B,C,D );
  pt aux = linesIntersection(A,B,C,D);
  return segContains(aux,A,B) && segContains(aux,C,D);
}

///CHANGE
lf distToSegment(pt p, pt A, pt B, pt &c) {
  lf u = dot( p-A , B-A ) / norm( B-A );
  if( u < -EPS ) { c = A; return dist( p , A ); }
  if( (u-1.0) > EPS ) { c = B; return dist( p, B ); }
  return distToLine(p,A,B,c);
}

// P[0] must be equal to P[n]
// Area is positive if the polygon is ccw
double signedArea(const vector<pt> &P) {
  double result = 0.0;
  for(int i = 0; i < (int)P.size()-1; i++) result += cross( P[i],P[i+1] );
  return result / 2.0;
}

double area(const vector<pt> &P) { return fabs(signedArea(P)); }

///CHANGE
// P[0] must be equal to P[n]
bool isConvex( const vector<pt> &P) {
  int sz = (int) P.size(); if(sz <= 3) return false;
  bool isL = ccw(P[0], P[1], P[2]) >= 0;
  for (int i = 1; i < sz-1; i++) {
    if( ( ccw(P[i], P[i+1], P[(i+2) == sz ? 1 : i+2]) >= 0 ) != isL) return false;
  }
  return true;
}

///CHANGE
// P[0] must be equal to P[n]     -    O(N)
// Return 1 for interior, 0 for boundary and -1 for exterior
int inPolygon(pt X, const vector<pt> &P) {
  const int n = P.size(); int cnt = 0;
  for (int i = 0; i < n-1; i++) {
    if( segContains(X, P[i], P[i+1]) ) return 0;
    if( y(P[i]) <= y(X) ) {
      if( y(P[i+1]) > y(X) )
        if( !(ccw( X, P[i], P[i+1]) >= 0) ) cnt++;
    }
    else if ( y(P[i+1]) <= y(X) ) {
      if( ccw( X, P[i], P[i+1]) >= 0 ) cnt--;
    }
  }
  if(cnt == 0) return -1;
  else return 1;
}

///CHANGE
//CONVEX HULL Monotone Chain O( N Log N )
bool cmpPt( pt A, pt B ) {
  if( !same( x(A), x(B) ) ) return x(A) < x(B);
  return y(A) < y(B);
}

int turn(pt A, pt B, pt C) {
  lf r = cross(B-A, C-A);
  if( same( r, 0.0 ) ) return 0;
  if( r > EPS ) return 1;
  return -1;
}

// Return CH in ccw order starting at leftmost - downmost x
// Doesn't return P[ n ] = P[ 0 ]
vector<pt> CH( vector<pt> &P ) {
  if ( P.size() == 1 ) return P;
  const int n = P.size();
  sort ( P.begin(), P.end(), cmpPt );
  vector<pt> up;
  up.push_back(P[0]); up.push_back(P[1]);
  vector<pt> dn;
  dn.push_back(P[0]); dn.push_back(P[1]);
  for ( int i = 2; i < n; ++i ) {
    // If collineal points are not needed, is >= and <=, otherwise > and <
    while ( up.size() > 1 && turn(up[up.size()-2],up.back(),P[i]) >= 0 )
      up.pop_back();
    while ( dn.size() > 1 && turn(dn[dn.size()-2],dn.back(),P[i]) <= 0 )
      dn.pop_back();
    up.push_back(P[i]);
    dn.push_back(P[i]);
  }
  for (int i = (int) up.size() - 2; i >= 1; i--) dn.push_back(up[i]);
  return dn;
}

///NEW
lf diameter( const vector<pt> &P, pt &A, pt &B ) {
  lf ans = -oo, d;
  int lo = 0, hi = 0;
  int sz = int(P.size());
  for( int i = 0, j = 0; i < sz; ++i ) {
    while( dist( P[i], P[j] )+EPS < dist( P[i], P[ (j+1)%sz ] )  ) {
      j = (j+1)%sz;
    }
    d = dist( P[i], P[j] );
    if( ans+EPS < d ) {
      ans = d;
      lo = i; hi = j;
    }
  }
  A = P[lo]; B = P[hi];
  return ans;
}

/// --------------------------------------------

//Returns the Polygon to the left of AB (counter clockwise)
vector<pt> cutPolygon (pt A, pt B, const vector<pt> &P) {
  vector<pt> Q;
  for (int i = 0; i < (int)P.size(); i++) {
    double left1 = cross( B-A , P[i]-A ), left2 = 0;
    if(i != (int)P.size()-1) left2 = cross( B-A , P[i+1]-A );
    if(left1 > -EPS) Q.push_back(P[i]);
    if( left1 * left2 < -EPS ) Q.push_back( linesIntersection(P[i], P[i+1], A, B) );
  }
  if (!Q.empty() && !samePt(Q.back(), Q.front()) ) Q.push_back(Q.front());
  return Q;
}

int n,m;

int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(10);
  cin >> n;
  vector<pt> P(n+1);
  for( int i = 0; i < n; ++i ) {
    cin >> P[i];
  }
  P[n] = P[0];
  cin >> m;
  for( int i = 0; i < m; ++i ) {
    pt A,B; cin >> A >> B;
    cout << double( area( cutPolygon( A, B, P ) ) ) << '\n';
  }
}