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>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops", \
                     "omit-frame-pointer", "inline")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,sse4.1,sse4.2,popcnt,abm,mmx,avx,avx2,fma,tune=native")
#pragma GCC option("arch=native", "no-zero-upper")
using namespace std;
const long long mod = (long long)(1e9 + 7);
const long double eps = (1e-9);
const long long oo = (long long)(1e18);
struct line {
  complex<long double> p, q;
};
double dot(complex<long double> a, complex<long double> b) {
  return real(conj(a) * b);
}
double cross(complex<long double> a, complex<long double> b) {
  return imag(conj(a) * b);
}
double area2(complex<long double> a, complex<long double> b,
             complex<long double> c) {
  return cross(b - a, c - a);
}
namespace std {
bool operator<(complex<long double> a, complex<long double> b) {
  if (a.real() != b.real()) return a.real() < b.real();
  return a.imag() < b.imag();
}
}  // namespace std
vector<complex<long double> > convex_hull(vector<complex<long double> > &P) {
  int n = P.size(), k = 0;
  vector<complex<long double> > h(2 * n);
  sort(P.begin(), P.end());
  for (int i = 0; i < n; h[k++] = P[i++])
    while (k >= 2 && area2(h[k - 2], h[k - 1], P[i]) < 0) --k;
  for (int i = n - 2, t = k + 1; i >= 0; h[k++] = P[i--])
    while (k >= t && area2(h[k - 2], h[k - 1], P[i]) < 0) --k;
  return vector<complex<long double> >(h.begin(), h.begin() + k - (k > 1));
}
complex<long double> operator*(long double x, complex<long double> a) {
  return complex<long double>(a.real() * x, a.imag() * x);
}
complex<long double> crosspoint(const line &l, const line &m) {
  double A = cross(l.q - l.p, m.q - m.p);
  double B = cross(l.q - l.p, l.q - m.p);
  if (abs(A) < eps && abs(B) < eps) return m.p;
  if (abs(A) < eps) assert(false);
  return m.p + B / A * (m.q - m.p);
}
complex<long double> read() {
  int x, y;
  cin >> x >> y;
  return complex<long double>(x, y);
}
void print(complex<long double> p) {
  cout << p.real() << ' ' << p.imag() << '\n';
}
long double solve(vector<complex<long double> > ps) {
  long double cnt = 1e4;
  int n = ps.size();
  for (int j = n - 1; j >= 0; j--) ps[j] = ps[j] - ps[0];
  for (int j = 1; j < n; j++) ps[j] = (cnt / (abs(ps[j]) * abs(ps[j]))) * ps[j];
  ps = convex_hull(ps);
  n = ps.size();
  bool ok = 1;
  for (int i = 0; i < n; i++)
    if (abs(ps[i]) < eps) ok = 0;
  if (!ok) return -1;
  ps.push_back(ps[0]);
  long double res = 0;
  for (int j = 0; j < n; j++) {
    complex<long double> p1 = ps[j];
    complex<long double> p2 = ps[j + 1];
    complex<long double> nor = complex<long double>(0, 1) * (p2 - p1);
    long double d = -dot(nor, p1);
    long double dist = abs(d) / abs(nor);
    res = max(res, cnt / (2 * dist));
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<complex<long double> > vr(n);
  vector<complex<long double> > vb(m);
  for (int i = 0; i < n; i++) vr[i] = read();
  for (int i = 0; i < m; i++) vb[i] = read();
  long double res = 0;
  for (int i = 0; i < m; i++) {
    vector<complex<long double> > cr;
    vector<complex<long double> > cb;
    for (int j = 0; j < n; j++) cr.push_back(vr[j] - vb[i]);
    for (int j = 0; j < m; j++) cb.push_back(vb[j] - vb[i]);
    long double cnt = 1e4;
    for (int j = 0; j < n; j++)
      cr[j] = (cnt / (abs(cr[j]) * abs(cr[j]))) * cr[j];
    for (int j = 0; j < m; j++) {
      if (i == j) continue;
      cb[j] = (cnt / (abs(cb[j]) * abs(cb[j]))) * cb[j];
    }
    cr = convex_hull(cr);
    cb = convex_hull(cb);
    cb.push_back(cb[0]);
    int lb = cb.size() - 1;
    int lr = cr.size();
    int best = 0;
    for (int j = 0; j < lb; j++) {
      complex<long double> p1 = cb[j];
      complex<long double> p2 = cb[j + 1];
      complex<long double> nor = complex<long double>(0, 1) * (p2 - p1);
      if (j == 0) {
        for (int k = 0; k < lr; k++)
          if (dot(nor, cr[k]) < dot(nor, cr[best])) best = k;
      } else {
        while (dot(nor, cr[best]) > dot(nor, cr[(best + 1) % lr]) + eps)
          best = (best + 1) % lr;
      }
      if (dot(nor, cr[best]) > dot(nor, p1) + eps) continue;
      long double d = -dot(nor, p1);
      long double dist = abs(d) / abs(nor);
      if (dist == 0)
        res = -1;
      else if (res != -1)
        res = max(res, cnt / (2 * dist));
    }
  }
  for (int i = 0; i < n; i++) {
    vector<complex<long double> > to_solve;
    to_solve.push_back(vr[i]);
    for (int j = 0; j < m; j++) to_solve.push_back(vb[j]);
    long double x = solve(to_solve);
    if (x == -1)
      res = -1;
    else if (res != -1)
      res = max(res, x);
  }
  cout.precision(10);
  cout << fixed;
  if (res == -1)
    cout << (-1) << '\n';
  else
    cout << (res) << '\n';
  return 0;
}
