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 N = 1054, INF = 2003731;
const long double eps = 2e-9l, pi2 = 6.283185307179586476925286766559005768l;
struct vec2 {
  long double x, y;
  vec2(long double x0 = 0.0, long double y0 = 0.0) : x(x0), y(y0) {}
  vec2 *read() {
    scanf("%Lf%Lf", &x, &y);
    return this;
  }
  inline vec2 operator-(const vec2 &B) const { return vec2(x - B.x, y - B.y); }
  inline long double norm2() const { return x * x + y * y; }
  inline long double norm() const { return sqrtl(x * x + y * y); }
  inline bool operator==(const vec2 &B) const { return x == B.x && y == B.y; }
};
int Re, Bl;
vec2 r[N], b[N];
std::pair<long double, int> D[N * 20];
inline bool intersection(const vec2 O1, const vec2 O2, long double r,
                         long double &beg, long double &end) {
  vec2 O12 = O2 - O1;
  long double d = O12.norm(), Cos = .5 * d / r, sAng = acosl(Cos),
              iAng = atan2l(O12.y, O12.x);
  return !(O1 == O2) && -1. <= Cos && Cos <= 1.
             ? (iAng < 0. ? iAng += pi2 : 0.,
                (beg = iAng - sAng) < 0. ? beg += pi2 : 0.,
                (end = iAng + sAng) > pi2 ? end -= pi2 : 0., true)
             : false;
}
int check(const vec2 O, long double radius, bool cover_red = false) {
  int i, s = 0, cnt = 0;
  long double il, ir;
  if (cover_red) D[cnt++] = std::pair<long double, int>(0, 1);
  for (i = 0; i < Re; ++i)
    if (intersection(O, r[i], radius, il, ir)) {
      if (il <= ir)
        D[cnt++] = std::pair<long double, int>(il, 1),
        D[cnt++] = std::pair<long double, int>(ir, -1);
      else
        D[cnt++] = std::pair<long double, int>(0, 1),
        D[cnt++] = std::pair<long double, int>(ir, -1),
        D[cnt++] = std::pair<long double, int>(il, 1);
    }
  for (i = 0; i < Bl; ++i)
    if (intersection(O, b[i], radius, il, ir)) {
      if (il <= ir)
        D[cnt++] = std::pair<long double, int>(il, -INF),
        D[cnt++] = std::pair<long double, int>(ir, INF);
      else
        D[cnt++] = std::pair<long double, int>(0, -INF),
        D[cnt++] = std::pair<long double, int>(ir, INF),
        D[cnt++] = std::pair<long double, int>(il, -INF);
    }
  std::sort(D, D + cnt);
  for (i = 0; i < cnt; ++i)
    if ((s += D[i].second) > 0) return true;
  return false;
}
int main() {
  int i;
  long double L = 1., R, M;
  char *_ = new char;
  std::mt19937 gen(time(NULL) + (unsigned long)_);
  delete _;
  scanf("%d%d", &Re, &Bl);
  for (i = 0; i < Re; ++i) r[i].read();
  for (i = 0; i < Bl; ++i) b[i].read();
  std::shuffle(r, r + Re, gen), std::shuffle(b, b + Bl, gen);
  for (i = 0; i < Re; ++i)
    if (check(r[i], L, true)) {
      for (R = 1e9; L + eps < R;
           check(r[i], M = (L + R) * .5, true) ? (L = M) : (R = M))
        ;
      L = (L + R) * .5;
    }
  for (i = 0; i < Bl; ++i)
    if (check(b[i], L)) {
      for (R = 1e9; L + eps < R;
           check(b[i], M = (L + R) * .5) ? (L = M) : (R = M))
        ;
      L = (L + R) * .5;
    }
  printf("%.12Lg\n", L > 1e9 - .3 ? -1. : L);
  return 0;
}
