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;
const double eps = 1e-6;
const double pi = acos(-1);
int sign(double k) {
  if (k > eps)
    return 1;
  else if (k < -eps)
    return -1;
  return 0;
}
int cmp(double k1, double k2) { return sign(k1 - k2); }
struct point {
  double x, y;
  point operator+(const point &k1) const { return (point){k1.x + x, k1.y + y}; }
  point operator-(const point &k1) const { return (point){x - k1.x, y - k1.y}; }
  point operator*(double k1) const { return (point){x * k1, y * k1}; }
  point operator/(double k1) const { return (point){x / k1, y / k1}; }
  int operator==(const point &k1) const {
    return cmp(x, k1.x) == 0 && cmp(y, k1.y) == 0;
  }
  point turn(double k1) {
    return (point){x * cos(k1) - y * sin(k1), x * sin(k1) + y * cos(k1)};
  }
  bool operator<(const point k1) const {
    int a = cmp(x, k1.x);
    if (a == -1)
      return 1;
    else if (a == 1)
      return 0;
    else
      return cmp(y, k1.y) == -1;
  }
  double abs() { return sqrt(x * x + y * y); }
  double abs2() { return x * x + y * y; }
  double dis(point k1) { return ((*this) - k1).abs(); }
  double getw() { return atan2(y, x); }
  int getP() const { return sign(y) == 1 || (sign(y) == 0 && sign(x) == -1); }
};
double cross(point k1, point k2) { return k1.x * k2.y - k1.y * k2.x; }
double dot(point k1, point k2) { return k1.x * k2.x + k1.y * k2.y; }
double slove(double S) {
  if (S < 0)
    S += 2 * pi;
  else if (S >= 2 * pi)
    S -= 2 * pi;
  return S;
}
point r[1005], b[1005];
vector<point> p;
int n, m;
vector<pair<double, int>> g;
bool check(int id, double R) {
  g.clear();
  point o = p[id];
  for (int i = 0; i < n + m; i++) {
    if (i == id) continue;
    point tmp = p[i];
    double d = tmp.dis(o);
    if (cmp(d, R * 2) >= 0) continue;
    double alf = acos(d / 2 / R);
    double delta = atan2(tmp.y - o.y, tmp.x - o.x);
    double l = slove(delta - alf), r = slove(delta + alf);
    if (l < r) {
      g.push_back(make_pair(l, i < n ? 2 : 1));
      g.push_back(make_pair(r, -(i < n ? 2 : 1)));
    } else {
      g.push_back(make_pair(l, i < n ? 2 : 1));
      g.push_back(make_pair(2 * pi, -(i < n ? 2 : 1)));
      g.push_back(make_pair(0, i < n ? 2 : 1));
      g.push_back(make_pair(r, -(i < n ? 2 : 1)));
    }
  }
  if (id < n)
    g.push_back(make_pair(0, id < n ? 2 : 1)),
        g.push_back(make_pair(2 * pi, -(id < n ? 2 : 1)));
  sort(g.begin(), g.end());
  int rn = 0, bn = 0;
  for (int i = 0; i < g.size(); i++) {
    if ((i == g.size() - 1 || g[i].first != g[i + 1].first) && rn > 0 &&
        bn <= 0)
      return true;
    if (abs(g[i].second) == 1)
      bn += g[i].second;
    else
      rn += g[i].second / 2;
  }
  return false;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++)
    scanf("%lf%lf", &r[i].x, &r[i].y), p.push_back(r[i]);
  for (int i = 0; i < m; i++)
    scanf("%lf%lf", &b[i].x, &b[i].y), p.push_back(b[i]);
  bool f = 0;
  for (int i = 0; i < n + m; i++)
    if (check(i, 1e15)) return 0 * printf("-1\n");
  double ans = 0;
  for (int i = 0; i < n + m; i++) {
    double l = ans, r = 1e9;
    if (check(i, ans)) {
      while (l + eps < r) {
        double mid = (l + r) / 2;
        if (check(i, mid))
          l = mid;
        else
          r = mid;
      }
      ans = l;
    }
  }
  printf("%.11f\n", ans);
}
