Problem: Given simple (without self-intersections) n-gon. It is not necessary convex. Also you are given m lines. For each line find the length of common part of the line and the n-gon.

The boundary of n-gon belongs to polygon. It is possible that n-gon contains 180-degree angles.

Input

The first line contains integers n and m (3 ≤ n ≤ 1000;1 ≤ m ≤ 100). The following n lines contain coordinates of polygon vertices (in clockwise or counterclockwise direction). All vertices are distinct.

The following m lines contain line descriptions. Each of them contains two distict points of a line by their coordinates.

All given in the input coordinates are real numbers, given with at most two digits after decimal point. They do not exceed 105 by absolute values.

Output

Print m lines, the i-th line should contain the length of common part of the given n-gon and the i-th line. The answer will be considered correct if the absolute or relative error doesn't exceed 10 - 6.

Examples

Input

4 3
0 0
1 0
1 1
0 1
0 0 1 1
0 0 0 1
0 0 1 -1


Output

1.41421356237309514547
1.00000000000000000000
0.00000000000000000000

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
bool zero(double x) {
  if (fabs(x) <= 1e-8) return true;
  return false;
}
int sign(double x) {
  if (zero(x)) return 0;
  if (x > 0) return 1;
  return -1;
}
struct point {
  double x, y;
  double tht;
  point(double xx = 0.0, double yy = 0.0) {
    x = xx;
    y = yy;
  }
  void input() { scanf("%lf%lf", &x, &y); }
  point operator-(const point &b) { return point(x - b.x, y - b.y); }
  point operator+(const point &b) { return point(x + b.x, y + b.y); }
  double operator^(const point &b) { return x * b.y - y * b.x; }
  double operator*(const point &b) { return x * b.x + y * b.y; }
};
struct xishu {
  double a, b, c;
};
double dist(point p1, point p2) { return sqrt((p1 - p2) * (p1 - p2)); }
xishu getline(point x, point y) {
  struct xishu w;
  w.a = y.y - x.y;
  w.b = x.x - y.x;
  w.c = y.x * x.y - x.x * y.y;
  return w;
}
int dots_inline(point p1, point p2, point p3) {
  return zero((p2 - p1) ^ (p3 - p1));
}
int dot_online_in(point p, point l1, point l2) {
  return ((l1 - p) * (l2 - p)) <= 1e-8;
}
point getnode(xishu l1, xishu l2) {
  point CrossP;
  CrossP.x = (l1.c * l2.b - l1.b * l2.c) / (l1.a * l2.b - l1.b * l2.a) * (-1.0);
  CrossP.y = (l1.c * l2.a - l1.a * l2.c) / (l1.b * l2.a - l1.a * l2.b) * (-1.0);
  return CrossP;
}
point p[1005];
int n;
bool dir(point A, point B) {
  if (zero(A.x - B.x)) return A.y < B.y;
  return A.x < B.x;
}
double dist2(point A, point B) {
  double dis = dist(A, B);
  if (dir(A, B))
    ;
  else
    dis *= -1;
  return dis;
}
bool cross(point l1, point l2, point s1, point s2) {
  point AB = l2 - l1;
  point AC = s1 - l1;
  point AD = s2 - l1;
  point CD = s2 - s1;
  point vec1 = l2 - l1;
  point vec2 = s2 - s1;
  if (((vec1 ^ vec2) < 0 ? -(vec1 ^ vec2) : (vec1 ^ vec2)) < 1e-8) return false;
  if ((AB ^ AC) >= -1e-8 && (AB ^ AD) >= -1e-8) return false;
  if ((AB ^ AC) < -1e-8 && (AB ^ AD) < -1e-8) return false;
  return true;
}
bool collinear(point l1, point l2, point s1, point s2) {
  if (dots_inline(l1, l2, s1) && dots_inline(l1, l2, s2)) return true;
  return false;
}
double solve(point l1, point l2) {
  vector<double> crs;
  vector<pair<double, double> > seg;
  for (int i = 0; i < n; i++) {
    int j = i + 1;
    if (cross(l1, l2, p[i], p[j])) {
      point CROSS = getnode(getline(l1, l2), getline(p[i], p[j]));
      crs.push_back(dist2(l1, CROSS));
    }
    if (collinear(l1, l2, p[i], p[j])) {
      double d1 = dist2(l1, p[i]);
      double d2 = dist2(l1, p[j]);
      if (d1 > d2) swap(d1, d2);
      seg.push_back(make_pair(d1, d2));
    }
  }
  sort(crs.begin(), crs.end());
  for (int j = 0; j + 1 < crs.size(); j += 2)
    seg.push_back(make_pair(crs[j], crs[j + 1]));
  sort(seg.begin(), seg.end());
  double ans = 0.0;
  double last = -1e18;
  for (int i = 0; i < seg.size(); i++) {
    if (seg[i].first > last) last = seg[i].first;
    if (seg[i].second > last) {
      ans = ans + seg[i].second - last;
      last = seg[i].second;
    }
  }
  return ans;
}
int main() {
  int k;
  scanf("%d", &n);
  scanf("%d", &k);
  memset(p, 0, sizeof(p));
  for (int i = 0; i < n; i++) p[i].input();
  p[n] = p[0];
  point s1, s2;
  for (int i = 0; i < k; i++) {
    s1.input();
    s2.input();
    double ans = solve(s1, s2);
    printf("%.12f\n", ans);
  }
  return 0;
}