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;
const long double EPS = 1e-7;
inline bool eq(long double x, long double y) { return fabs(x - y) < EPS; }
inline bool le(long double x, long double y) { return x < y + EPS; }
inline bool lt(long double x, long double y) { return x + EPS < y; }
inline int sgn(long double x) { return eq(x, 0) ? 0 : x < 0 ? -1 : 1; }
struct point {
  long double x, y;
  point(long double x = 0, long double y = 0) : x(x), y(y) {}
  point operator+(const point& p) const { return point(x + p.x, y + p.y); }
  point operator-(const point& p) const { return point(x - p.x, y - p.y); }
  point operator*(long double s) const { return point(x * s, y * s); }
  point operator/(long double s) const { return point(x / s, y / s); }
  long double operator*(const point& p) const { return x * p.x + y * p.y; }
  long double operator%(const point& p) const { return x * p.y - y * p.x; }
  long double norm_sq() { return *this * *this; }
  long double norm() { return sqrt(*this * *this); }
  bool operator==(const point& p) const { return eq(x, p.x) && eq(y, p.y); }
  bool operator<(const point& p) const {
    return eq(y, p.y) ? lt(x, p.x) : lt(y, p.y);
  }
};
ostream& operator<<(ostream& o, const point& p) {
  return o << "(" << p.x << " " << p.y << ")";
}
int ccw(point a, point b, point c) { return sgn((b - a) % (c - b)); }
struct line {
  point a, ab;
  line(point a, point b) : a(a), ab(b - a) {}
  point b() { return a + ab; }
  bool areParallel(const line& l) const { return eq(ab % l.ab, 0); }
};
bool areIntersect(line l1, line l2, point& res) {
  if (l1.areParallel(l2)) return 0;
  long double ls = (l2.a - l1.a) % l2.ab, rs = l1.ab % l2.ab;
  res = l1.a + l1.ab * ls / rs;
  return 1;
}
bool areInterLineSeg(line l, point a, point b, point& res) {
  if (!areIntersect(l, line(a, b), res)) return 0;
  if (le((a - res) * (b - res), 0)) return 1;
  return 0;
}
int pointVsPolygon(point p, const vector<point>& vp) {
  int wn = 0, n = (int)vp.size() - 1;
  for (int i = 0; i < n; i++) {
    int cs = ccw(vp[i + 1], vp[i], p);
    if (cs == 0 && (vp[i + 1] - p) * (vp[i] - p) <= 0) return 0;
    if (vp[i].y <= p.y) {
      if (vp[i + 1].y > p.y && cs > 0) wn++;
    } else {
      if (vp[i + 1].y <= p.y && cs < 0) wn--;
    }
  }
  return wn == 0 ? 1 : -1;
}
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  vector<point> poly;
  for (int i = 0; i < n; ++i) {
    double x, y;
    scanf("%lf %lf", &x, &y);
    poly.push_back(point(x, y));
  }
  poly.push_back(poly.front());
  for (int i = 0; i < m; ++i) {
    double xa, xb, ya, yb;
    scanf("%lf %lf %lf %lf", &xa, &ya, &xb, &yb);
    point a(xa, ya), b(xb, yb);
    line l(a, b);
    vector<pair<point, pair<int, int> > > ints;
    point p;
    for (int j = 0; j < n; ++j) {
      if (!areInterLineSeg(l, poly[j], poly[j + 1], p)) continue;
      int pre = ccw(a, b, poly[j]), nxt = ccw(a, b, poly[j + 1]);
      ints.push_back({p, {pre, nxt}});
    }
    sort(ints.begin(), ints.end());
    long double ans = 0;
    int inside = -1, prev = 0;
    point last;
    for (int j = 0; j < ints.size(); ++j) {
      point c = (ints[j].first + ints[j + 1].first) / 2.0;
      if (inside >= 0) ans += (ints[j].first - last).norm();
      last = ints[j].first;
      if (inside == 0) {
        int cur = ((ints[j].second.first + ints[j].second.second));
        assert(ints[j].second.first == 0 || ints[j].second.second == 0);
        if (cur != 0) {
          if (cur == prev)
            inside = 1;
          else
            inside = -1;
        }
      } else if (ints[j].second.first * ints[j].second.second < 0) {
        if (inside >= 0)
          inside = -1;
        else
          inside = 1;
      } else {
        prev = (ints[j].second.first + ints[j].second.second) * inside;
        inside = 0;
      }
    }
    printf("%.20lf\n", (double)ans);
  }
  return 0;
}