#include <bits/stdc++.h>
using namespace std;
const double PI = (acos(-1.0));
struct Point {
  long long x, y;
  Point(long long ix = 0, long long iy = 0) : x(ix), y(iy) {}
  Point &operator=(const Point &b) {
    x = b.x;
    y = b.y;
    return *this;
  }
  Point &operator-=(const Point &b) {
    x -= b.x;
    y -= b.y;
    return *this;
  }
  Point operator-(const Point &b) {
    Point res = *this;
    res -= b;
    return res;
  }
  friend long long dot(const Point &a, const Point &b) {
    return a.x * b.x + a.y * b.y;
  }
  friend long long cross(const Point &a, const Point &b) {
    return a.x * b.y - b.x * a.y;
  }
  friend double dist(const Point &a) {
    return sqrt(double(a.x * a.x + a.y * a.y));
  }
  friend double dist(const Point &a, const Point &b) {
    Point c = a;
    c -= b;
    return dist(c);
  }
  bool operator<(const Point &b) {
    const Point &a = *this;
    bool ka = a.y < 0 || a.y == 0 && a.x < 0;
    bool kb = b.y < 0 || b.y == 0 && b.x < 0;
    return ka < kb || ka == kb && cross(a, b) > 0;
  }
};
double getAnsver(Point p[1002], Point line[2], int n) {
  vector<pair<double, int>> ans;
  double a = dist(line[1]);
  int i;
  i = (1);
  for (int _n = (int)(n + 1), _di = (i > _n) ? -1 : 1; i != _n; i += _di) {
    long long y0 = cross(p[i] - line[0], line[1]);
    long long y1 = cross(p[i + 1] - line[0], line[1]);
    long long y2 = cross(p[i - 1] - line[0], line[1]);
    if (y0 == 0) {
      if ((y1 == 0) != (y2 == 0)) {
        double res = dist(p[i], line[0]);
        int flag = (y1 + y2 > 0) ? 1 : -1;
        ans.push_back(make_pair(res, flag));
      } else if (y1 < 0 != y2 < 0) {
        double res = dist(p[i], line[0]);
        ans.push_back(make_pair(res, 0));
      }
    } else if (y0 > 0 && y1 < 0 || y0 < 0 && y1 > 0) {
      double a0 = cross(p[i + 1] - p[i], line[0] - p[i]);
      double ca0 = dot(p[i + 1] - p[i], line[0] - p[i]);
      double cy0 = dot(p[i] - line[0], line[1]);
      double bb = dot(p[i] - line[0], p[i] - line[0]);
      double res = bb * a * a0 / (a0 * cy0 + ca0 * y0);
      ans.push_back(make_pair(fabs(res), 0));
    }
  }
  double result = 0.;
  bool inPolygon = false;
  sort(ans.begin(), ans.end());
  i = (0);
  for (int _n = (int)(ans.size()), _di = (i > _n) ? -1 : 1; i != _n; i += _di) {
    if (inPolygon) {
      result += ans[i].first;
      inPolygon = false;
      if (ans[i].second && i + 1 < ans.size() && ans[i + 1].second) {
        result += ans[i + 1].first - ans[i].first;
        if (ans[i].second * ans[i + 1].second == 1) {
          result -= ans[i + 1].first;
          inPolygon = true;
        }
        i++;
      }
    } else {
      result -= ans[i].first;
      inPolygon = true;
      if (ans[i].second && i + 1 < ans.size() && ans[i + 1].second) {
        if (ans[i].second * ans[i + 1].second == 1) {
          result += ans[i + 1].first;
          inPolygon = false;
        }
        i++;
      }
    }
  }
  ans.clear();
  return result;
}
int main() {
  ios::sync_with_stdio(false);
  int n, m, i, j;
  double x, y;
  Point p[1002], line[2];
  cin >> n >> m;
  i = (0);
  for (int _n = (int)(n), _di = (i > _n) ? -1 : 1; i != _n; i += _di) {
    cin >> x >> y;
    p[i] = Point(x * 100, y * 100);
  }
  p[n] = p[0];
  p[n + 1] = p[1];
  j = (0);
  for (int _n = (int)(m), _di = (j > _n) ? -1 : 1; j != _n; j += _di) {
    cin >> x >> y;
    line[0] = Point(x * 100, y * 100);
    cin >> x >> y;
    line[1] = Point(x * 100, y * 100);
    line[1] -= line[0];
    long long step = 3e7 / max(abs(line[1].x), abs(line[1].y));
    line[0].x += step * line[1].x;
    line[0].y += step * line[1].y;
    cout << setprecision(16) << getAnsver(p, line, n) / 100 << endl;
  }
  return 0;
}

### Prompt

Please formulate a CPP solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const double PI = (acos(-1.0));
struct Point {
  long long x, y;
  Point(long long ix = 0, long long iy = 0) : x(ix), y(iy) {}
  Point &operator=(const Point &b) {
    x = b.x;
    y = b.y;
    return *this;
  }
  Point &operator-=(const Point &b) {
    x -= b.x;
    y -= b.y;
    return *this;
  }
  Point operator-(const Point &b) {
    Point res = *this;
    res -= b;
    return res;
  }
  friend long long dot(const Point &a, const Point &b) {
    return a.x * b.x + a.y * b.y;
  }
  friend long long cross(const Point &a, const Point &b) {
    return a.x * b.y - b.x * a.y;
  }
  friend double dist(const Point &a) {
    return sqrt(double(a.x * a.x + a.y * a.y));
  }
  friend double dist(const Point &a, const Point &b) {
    Point c = a;
    c -= b;
    return dist(c);
  }
  bool operator<(const Point &b) {
    const Point &a = *this;
    bool ka = a.y < 0 || a.y == 0 && a.x < 0;
    bool kb = b.y < 0 || b.y == 0 && b.x < 0;
    return ka < kb || ka == kb && cross(a, b) > 0;
  }
};
double getAnsver(Point p[1002], Point line[2], int n) {
  vector<pair<double, int>> ans;
  double a = dist(line[1]);
  int i;
  i = (1);
  for (int _n = (int)(n + 1), _di = (i > _n) ? -1 : 1; i != _n; i += _di) {
    long long y0 = cross(p[i] - line[0], line[1]);
    long long y1 = cross(p[i + 1] - line[0], line[1]);
    long long y2 = cross(p[i - 1] - line[0], line[1]);
    if (y0 == 0) {
      if ((y1 == 0) != (y2 == 0)) {
        double res = dist(p[i], line[0]);
        int flag = (y1 + y2 > 0) ? 1 : -1;
        ans.push_back(make_pair(res, flag));
      } else if (y1 < 0 != y2 < 0) {
        double res = dist(p[i], line[0]);
        ans.push_back(make_pair(res, 0));
      }
    } else if (y0 > 0 && y1 < 0 || y0 < 0 && y1 > 0) {
      double a0 = cross(p[i + 1] - p[i], line[0] - p[i]);
      double ca0 = dot(p[i + 1] - p[i], line[0] - p[i]);
      double cy0 = dot(p[i] - line[0], line[1]);
      double bb = dot(p[i] - line[0], p[i] - line[0]);
      double res = bb * a * a0 / (a0 * cy0 + ca0 * y0);
      ans.push_back(make_pair(fabs(res), 0));
    }
  }
  double result = 0.;
  bool inPolygon = false;
  sort(ans.begin(), ans.end());
  i = (0);
  for (int _n = (int)(ans.size()), _di = (i > _n) ? -1 : 1; i != _n; i += _di) {
    if (inPolygon) {
      result += ans[i].first;
      inPolygon = false;
      if (ans[i].second && i + 1 < ans.size() && ans[i + 1].second) {
        result += ans[i + 1].first - ans[i].first;
        if (ans[i].second * ans[i + 1].second == 1) {
          result -= ans[i + 1].first;
          inPolygon = true;
        }
        i++;
      }
    } else {
      result -= ans[i].first;
      inPolygon = true;
      if (ans[i].second && i + 1 < ans.size() && ans[i + 1].second) {
        if (ans[i].second * ans[i + 1].second == 1) {
          result += ans[i + 1].first;
          inPolygon = false;
        }
        i++;
      }
    }
  }
  ans.clear();
  return result;
}
int main() {
  ios::sync_with_stdio(false);
  int n, m, i, j;
  double x, y;
  Point p[1002], line[2];
  cin >> n >> m;
  i = (0);
  for (int _n = (int)(n), _di = (i > _n) ? -1 : 1; i != _n; i += _di) {
    cin >> x >> y;
    p[i] = Point(x * 100, y * 100);
  }
  p[n] = p[0];
  p[n + 1] = p[1];
  j = (0);
  for (int _n = (int)(m), _di = (j > _n) ? -1 : 1; j != _n; j += _di) {
    cin >> x >> y;
    line[0] = Point(x * 100, y * 100);
    cin >> x >> y;
    line[1] = Point(x * 100, y * 100);
    line[1] -= line[0];
    long long step = 3e7 / max(abs(line[1].x), abs(line[1].y));
    line[0].x += step * line[1].x;
    line[0].y += step * line[1].y;
    cout << setprecision(16) << getAnsver(p, line, n) / 100 << endl;
  }
  return 0;
}

```