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 long double EPS = 1e-9;
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) {
  long long n = P.size(), k = 0;
  vector<complex<long double>> h(2 * n);
  sort(P.begin(), P.end());
  for (long long i = 0; i < n; h[k++] = P[i++])
    while (k >= 2 && area2(h[k - 2], h[k - 1], P[i]) < EPS) --k;
  for (long long i = n - 2, t = k + 1; i >= 0; h[k++] = P[i--])
    while (k >= t && area2(h[k - 2], h[k - 1], P[i]) < EPS) --k;
  return vector<complex<long double>>(h.begin(), h.begin() + k - (k > 1));
}
vector<pair<complex<long double>, long long>> CH(
    vector<pair<complex<long double>, long long>> &P) {
  long long n = P.size(), k = 0;
  vector<pair<complex<long double>, long long>> h(2 * n);
  sort(P.begin(), P.end());
  for (long long i = 0; i < n; h[k++] = P[i++])
    while (k >= 2 && area2(h[k - 2].first, h[k - 1].first, P[i].first) < EPS)
      --k;
  for (long long i = n - 2, t = k + 1; i >= 0; h[k++] = P[i--])
    while (k >= t && area2(h[k - 2].first, h[k - 1].first, P[i].first) < EPS)
      --k;
  h.resize(k - (k > 1));
  return h;
}
vector<complex<long double>> convex_hull2(vector<complex<long double>> &P) {
  long long n = P.size(), k = 0;
  vector<complex<long double>> h(2 * n);
  sort(P.begin(), P.end());
  for (long long i = 0; i < n; h[k++] = P[i++])
    while (k >= 2 && area2(h[k - 2], h[k - 1], P[i]) < -EPS) --k;
  for (long long i = n - 2, t = k + 1; i >= 0; h[k++] = P[i--])
    while (k >= t && area2(h[k - 2], h[k - 1], P[i]) < -EPS) --k;
  return vector<complex<long double>>(h.begin(), h.begin() + k - (k > 1));
}
long double dist(complex<long double> p1, complex<long double> p2) {
  long double xx = p1.real() - p2.real();
  long double yy = p1.imag() - p2.imag();
  return sqrtl(xx * xx + yy * yy);
}
long double radio(complex<long double> p1, complex<long double> p2,
                  complex<long double> p3) {
  long double d1 = dist(p1, p2);
  long double d2 = dist(p1, p3);
  long double d3 = dist(p2, p3);
  long double S = abs(area2(p1, p2, p3)) / 2.0;
  if (S < EPS) return -1e18;
  return d1 * d2 * d3 / S / 4.0;
}
complex<long double> invert(complex<long double> p1, long double rad,
                            complex<long double> p2) {
  long double d1 = dist(p1, p2);
  long double d2 = rad * rad / d1;
  long double prop = d2 / d1;
  p2 -= p1;
  p2 *= prop;
  p2 += p1;
  return p2;
}
const long double RADIO = 100;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout << setprecision(10) << fixed << '\n';
  long long N, M;
  cin >> N >> M;
  vector<complex<long double>> v1, v2;
  for (long long i = 1; i <= N; i++) {
    long double x, y;
    cin >> x >> y;
    v1.push_back({x, y});
  }
  for (long long i = 1; i <= M; i++) {
    long double x, y;
    cin >> x >> y;
    v2.push_back({x, y});
  }
  swap(v1, v2);
  swap(N, M);
  if (N == 1) {
    cout << -1 << '\n';
    return 0;
  }
  long double res = -1e18;
  bool flag = true;
  for (long long i = 0; i < N; i++) {
    vector<complex<long double>> vect;
    vector<complex<long double>> vect2;
    for (long long j = 0; j < N; j++) {
      if (i == j) {
        vect.push_back(v1[i]);
      } else {
        vect.push_back(invert(v1[i], RADIO, v1[j]));
      }
    }
    set<pair<complex<long double>, complex<long double>>> st;
    set<pair<complex<long double>, complex<long double>>> st2;
    vect = convex_hull(vect);
    vect2 = vect;
    for (long long j = 0; j < M; j++) {
      vect2.push_back(invert(v1[i], RADIO, v2[j]));
    }
    vect2 = convex_hull2(vect2);
    for (long long j = 0; j < vect2.size(); j++) {
      complex<long double> p1 = vect2[j];
      complex<long double> p2 = vect2[(j + 1) % vect2.size()];
      if (p2 < p1) swap(p1, p2);
      st.insert({p1, p2});
    }
    auto vect3 = convex_hull(vect2);
    for (long long j = 0; j < vect3.size(); j++) {
      complex<long double> p1 = vect3[j];
      complex<long double> p2 = vect3[(j + 1) % vect3.size()];
      if (p2 < p1) swap(p1, p2);
      st2.insert({p1, p2});
    }
    for (long long j = 0; j < vect.size(); j++) {
      complex<long double> p1 = vect[j];
      complex<long double> p2 = vect[(j + 1) % vect.size()];
      if (p2 < p1) swap(p1, p2);
      if (st2.find({p1, p2}) == st2.end()) {
        if (p1 == v1[i] || p2 == v1[i]) {
          flag = false;
        }
      }
      if (st.find({p1, p2}) == st.end()) {
        if (p1 == v1[i] || p2 == v1[i]) {
        } else {
          p1 = invert(v1[i], RADIO, p1);
          p2 = invert(v1[i], RADIO, p2);
          res = max(res, radio(v1[i], p1, p2));
        }
      }
    }
    vector<pair<complex<long double>, long long>> vvv;
    for (long long j = 0; j < N; j++) {
      if (i == j) {
        vvv.push_back({v1[i], 1});
      } else {
        vvv.push_back({invert(v1[i], RADIO, v1[j]), 1});
      }
    }
    for (long long j = 0; j < M; j++) {
      vvv.push_back({invert(v1[i], RADIO, v2[j]), 2});
    }
    vvv = CH(vvv);
    for (long long j = 0; j < vvv.size(); j++) {
      long long k1 = vvv[j].second;
      long long k2 = vvv[(j + 1) % vvv.size()].second;
      if (k1 == k2) continue;
      complex<long double> p1 = vvv[j].first;
      complex<long double> p2 = vvv[(j + 1) % vvv.size()].first;
      if (p1 == v1[i] || p2 == v1[i]) continue;
      p1 = invert(v1[i], RADIO, p1);
      p2 = invert(v1[i], RADIO, p2);
      res = max(res, radio(v1[i], p1, p2));
    }
  }
  if (flag && res > 0) {
    cout << res << '\n';
  } else {
    cout << -1 << '\n';
  }
  return 0;
}
