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>
const int MAXN = 2010;
const long double eps = 1e-6;
bool le(long double a, long double b) { return a + eps < b; }
bool ge(long double a, long double b) { return a - eps > b; }
bool leq(long double a, long double b) { return a - eps <= b; }
bool geq(long double a, long double b) { return a + eps >= b; }
bool eq(long double a, long double b) { return a <= b + eps && b <= a + eps; }
struct vec {
  long double x, y;
  vec() { x = y = 0; }
  vec(long double a, long double b) { x = a, y = b; }
  long double operator^(vec b) const { return x * b.y - y * b.x; }
  long double operator*(vec b) const { return x * b.x + y * b.y; }
  long double norm2() const { return x * x + y * y; }
  long double norm() const { return sqrt(x * x + y * y); }
  vec operator-(vec b) const { return vec(x - b.x, y - b.y); }
  vec operator+(vec b) const { return vec(x + b.x, y + b.y); }
  vec operator*(long double k) const { return vec(x * k, y * k); }
  bool operator<(const vec& b) const { return x == b.x ? y < b.y : x < b.x; }
  friend std::istream& operator>>(std::istream& in, vec& x) {
    in >> x.x >> x.y;
    return in;
  }
  friend std::ostream& operator<<(std::ostream& out, vec& x) {
    out << x.x << ' ' << x.y;
    return out;
  }
};
inline long double cross(vec a, vec b, vec c) { return (b - a) ^ (c - a); }
int n, m;
vec A[MAXN];
std::vector<std::pair<long double, int> > V;
long double nil = -123456789;
const long double PI = acosl(-1);
long double red(long double x) { return fmodl(x + PI * 2 * 5, PI * 2); }
std::pair<long double, long double> calc(vec x, vec y, long double r) {
  if (ge((x = y - x).norm2(), r * r * 4))
    return std::pair<long double, long double>(nil, nil);
  long double dis = x.norm();
  long double delta = acosl(dis / 2 / r);
  long double deg = atan2l(x.y, x.x);
  return std::pair<long double, long double>(red(deg - delta),
                                             red(deg + delta));
}
bool solve(int idx, long double R) {
  V.clear();
  if (idx <= n) V.emplace_back((long double)0, 1);
  for (int i = 1; i <= n + m; ++i)
    if (i != idx) {
      std::pair<long double, long double> r = calc(A[idx], A[i], R);
      if (r.first == nil) continue;
      const int v = i <= n ? 1 : -1e6;
      if (r.first > r.second) V.emplace_back((long double)0, v);
      V.emplace_back(r.first, v);
      V.emplace_back(r.second, -v);
    }
  std::sort(V.begin(), V.end());
  int now = 0;
  for (auto t : V)
    if ((now += t.second) > 0) return true;
  return false;
}
const long double ma = 1e9;
int main() {
  std::ios_base::sync_with_stdio(false), std::cin.tie(0);
  std::cin >> n >> m;
  for (int i = 1; i <= n + m; ++i) std::cin >> A[i];
  long double l = 0.5;
  for (int i = 1; i <= n + m; ++i)
    if (solve(i, l)) {
      long double r = ma, mid;
      while (le(l, r)) solve(i, mid = (l + r) / 2) ? l = mid : r = mid;
    }
  if (l > ma - 1.2243)
    std::cout << -1 << std::endl;
  else
    std::cout << std::fixed << std::setprecision(10) << l << std::endl;
  return 0;
}
