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;
bool debug = 0;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
string direc = "RDLU";
long long ln, lk, lm;
void etp(bool f = 0) {
  puts(f ? "yes" : "no");
  exit(0);
}
void addmod(int &x, int y, int mod = 1000000007) {
  x += y;
  if (x >= mod) x -= mod;
  assert(x >= 0 && x < mod);
}
void et() {
  puts("-1");
  exit(0);
}
struct Point {
  long double x, y;
  Point() {}
  Point(long double x, long double y) : x(x), y(y) {}
  long double abs() const { return hypot(x, y); }
  long double arg() const { return atan2(y, x); }
  Point operator*(double o) const { return Point(x * o, y * o); }
  Point operator+(const Point &o) const { return Point(x + o.x, y + o.y); }
  Point operator-(const Point &o) const { return Point(x - o.x, y - o.y); }
  bool operator<(const Point &o) const {
    return x < o.x - 1e-6 || (x < o.x + 1e-6 && y < o.y - 1e-6);
  }
  friend bool operator==(const Point &r1, const Point &r2) {
    return r1.x == r2.x && r1.y == r2.y;
  }
  Point scale(double o) const { return *this * (o / abs()); }
  Point rotY() const { return Point(-y, x); }
  Point rotX() const { return Point(y, -x); }
  long double cross(Point b) const { return x * b.y - b.x * y; }
  long double dot(Point b) const { return x * b.x + y * b.y; }
  long double disToSeg(Point &a, Point &b) {
    return fabs(((*this) - a).cross(b - a) / (b - a).abs());
  }
  void readin() { scanf("%Lf%Lf", &x, &y); }
} p[2135];
void fm(long double &x) {
  if (x <= -acos(-1)) x += acos(-1) * 2;
  if (x > acos(-1)) x -= acos(-1) * 2;
}
bool ck(int tar, long double r) {
  vector<pair<long double, int>> vp;
  for (int(i) = 1; (i) <= (int)(n + m); (i)++)
    if (i != tar) {
      long double dis = (p[i] - p[tar]).abs();
      if (dis > r + r) continue;
      long double arc = (p[i] - p[tar]).arg();
      long double the = acos(dis / 2 / r);
      long double a2 = arc + the, a1 = arc - the;
      fm(a1);
      fm(a2);
      if (a1 <= a2) {
        vp.push_back({a1, i <= n ? 1 : -(1 << 30)});
        vp.push_back({a2, i <= n ? -1 : (1 << 30)});
      } else {
        vp.push_back({-acos(-1), i <= n ? 1 : -(1 << 30)});
        vp.push_back({a2, i <= n ? -1 : (1 << 30)});
        vp.push_back({a1, i <= n ? 1 : -(1 << 30)});
        vp.push_back({acos(-1), i <= n ? -1 : (1 << 30)});
      }
    }
  if (tar <= n) {
    vp.push_back({-acos(-1), 1});
    vp.push_back({acos(-1), -1});
  }
  sort(vp.begin(), vp.end());
  long long cc = 0;
  for (auto p : vp) {
    cc += p.second;
    if (cc > 0) return 1;
  }
  return 0;
}
void fmain() {
  scanf("%d%d", &n, &m);
  for (int(i) = 1; (i) <= (int)(n + m); (i)++) p[i].readin();
  long double ans = 0;
  for (int(i) = 1; (i) <= (int)(n + m); (i)++)
    if (ck(i, ans)) {
      if (ck(i, 1e11)) et();
      long double l = ans, r = 1e11;
      while (r - l > 1e-6) {
        long double mid = (l + r) / 2;
        if (ck(i, mid))
          l = mid;
        else
          r = mid;
      }
      ans = l;
    }
  printf("%.10Lf\n", ans);
}
int main() {
  int t = 1;
  for (int(i) = 1; (i) <= (int)(t); (i)++) {
    fmain();
  }
  return 0;
}
