Hongcow really likes the color red. Hongcow doesn't like the color blue.

Hongcow is standing in an infinite field where there are n red points and m blue points.

Hongcow wants to draw a circle in the field such that this circle contains at least one red point, and no blue points. Points that line exactly on the boundary of the circle can be counted as either inside or outside.

Compute the radius of the largest circle that satisfies this condition. If this circle can have arbitrarily large size, print  - 1. Otherwise, your answer will be accepted if it has relative or absolute error at most 10 - 4.

Input

The first line of the input will contain two integers n, m (1 ≤ n, m ≤ 1, 000).

The next n lines will contain two integers xi, yi (1 ≤ xi, yi ≤ 104). This denotes the coordinates of a red point.

The next m lines will contain two integers xi, yi (1 ≤ xi, yi ≤ 104). This denotes the coordinates of a blue point.

No two points will have the same coordinates.

Output

Print  - 1 if the circle can have arbitrary size. Otherwise, print a floating point number representing the largest radius circle that satisfies the conditions. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 4.

Namely, let's assume that your answer is a and the answer of the jury is b. The checker program will consider your answer correct if <image>.

Examples

Input

2 5
2 3
3 4
1 1
1 4
4 2
4 7
2 5


Output

3.5355338827


Input

1 6
3 3
1 5
5 4
2 1
3 4
4 2
1 3


Output

1.5811388195


Input

2 2
2 2
3 3
1 1
4 4


Output

-1

Note

This is a picture of the first sample 

<image>

This is a picture of the second sample 

<image>
#include <bits/stdc++.h>
using namespace std;
template <class T, class S>
ostream& operator<<(ostream& os, const pair<T, S>& v) {
  return os << "(" << v.first << ", " << v.second << ")";
}
template <class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "[";
  for (int i = int(0); i <= int((static_cast<int>((v).size())) - 1); ++i) {
    if (i) os << ", ";
    os << v[i];
  }
  return os << "]";
}
template <class T>
bool setmax(T& _a, T _b) {
  if (_b > _a) {
    _a = _b;
    return true;
  }
  return false;
}
template <class T>
bool setmin(T& _a, T _b) {
  if (_b < _a) {
    _a = _b;
    return true;
  }
  return false;
}
template <class T>
T gcd(T _a, T _b) {
  return _b == 0 ? _a : gcd(_b, _a % _b);
}
const long double PI = acosl(-1);
int sgn(long double x, long double e = 1e-15) { return x < -e ? -1 : x > e; }
struct Point {
  long double x, y;
  int label = 0;
  Point() : x(0), y(0) {}
  Point(long double _x, long double _y) : 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 b) const { return Point(x * b, y * b); }
  Point operator/(long double b) const { return Point(x / b, y / b); }
  bool operator==(Point p) const {
    return sgn(x - p.x) == 0 && sgn(y - p.y) == 0;
  }
  long double sqr_abs() const { return x * x + y * y; }
  long double abs() const { return sqrt(sqr_abs()); }
  long double ang() const { return atan2l(y, x); }
  friend long double mult(Point p1, Point p2) {
    return p1.x * p2.y - p1.y * p2.x;
  }
  friend long double mult(Point p0, Point p1, Point p2) {
    return mult(p1 - p0, p2 - p0);
  }
  friend long double dot(Point p1, Point p2) {
    return p1.x * p2.x + p1.y * p2.y;
  }
  friend long double dot(Point p0, Point p1, Point p2) {
    return dot(p1 - p0, p2 - p0);
  }
  void print() const { printf("%.2lf %.2lf", (double)x, (double)y); }
};
Point read_point() {
  double u, v;
  scanf("%lf%lf", &u, &v);
  return Point(u, v);
}
vector<Point> ps;
long double normalize(long double ang) {
  if (ang < -PI) ang += PI * 2;
  if (ang > +PI) ang -= PI * 2;
  return ang;
}
bool check(Point center, long double r) {
  struct Event {
    long double t;
    int label;
    int stub;
    bool operator<(const Event& e) const {
      int s = sgn(t - e.t);
      if (s != 0) return s < 0;
      if (stub != e.stub) return stub < 0;
      if (stub > 0) {
        return label > e.label;
      } else {
        return label < e.label;
      }
    }
  };
  vector<Event> es;
  for (const auto& p : ps) {
    if (p == center) continue;
    if (center.label && p.label) continue;
    long double d = (p - center).abs();
    if (sgn(d - r * 2) > 0) continue;
    long double dir = (p - center).ang();
    long double phi = acosl(d / 2 / r);
    long double st = normalize(dir - phi), en = normalize(dir + phi);
    if (sgn(st - en) <= 0) {
      es.push_back({st, p.label, 1});
      es.push_back({en, p.label, -1});
    } else {
      es.push_back({st, p.label, 1});
      es.push_back({PI + 1, p.label, -1});
      es.push_back({-PI - 1, p.label, 1});
      es.push_back({en, p.label, -1});
    }
  }
  if (center.label) {
    es.push_back({-PI - 1, 1, 1});
    es.push_back({0, 1, -1});
    es.push_back({0, 1, 1});
    es.push_back({PI + 1, 1, -1});
  }
  sort((es).begin(), (es).end());
  int cnt[2] = {0, 0};
  for (const auto& e : es) {
    cnt[e.label] += e.stub;
    if (sgn(e.t + PI) >= 0 && sgn(e.t - PI) <= 0) {
      if (cnt[1] > 0 && cnt[0] == 0) {
        return true;
      }
    }
  }
  return false;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = int(0); i <= int((n)-1); ++i) {
    ps.push_back(read_point());
    ps.back().label = 1;
  }
  for (int i = int(0); i <= int((m)-1); ++i) {
    ps.push_back(read_point());
  }
  for (const auto& p : ps) {
    if (p.label && check(p, 1e15)) {
      printf("-1\n");
      return 0;
    }
  }
  long double ans = 0.1;
  for (int label = int(1); label >= int(0); --label) {
    for (const auto& p : ps) {
      if (p.label == label) {
        if (!check(p, ans)) continue;
        long double l = ans, r = ans * 2;
        while (check(p, r)) r *= 2;
        for (int i = int(0); i <= int((70) - 1); ++i) {
          long double mid = (l + r) / 2;
          if (check(p, mid))
            l = mid;
          else
            r = mid;
        }
        ans = l;
      }
    }
  }
  printf("%.9lf\n", (double)ans);
  return 0;
}
