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;
int T, n, m, k, q, t;
const long double eps = 1e-12, P = acos(-1.0);
struct point {
  long double x, y;
  point(long double xx = 0, long double yy = 0) { x = xx, y = yy; }
  point operator-(const point &b) const { return point(x - b.x, y - b.y); }
  point operator+(const point &b) const { return point(x + b.x, y + b.y); }
  point operator*(const long double &b) const { return point(x * b, y * b); }
  long double operator*(const point &b) const { return x * b.x + y * b.y; }
} p[2010];
struct node {
  long double w;
  int id;
  node(long double ww = 0, int iid = 0) { w = ww, id = iid; }
  bool operator<(const node &b) const { return w < b.w; }
} st[8010];
long double dis(point a, point b) {
  return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
long double angle(point a, point b, long double r) {
  return acos(dis(a, b) / (2 * r));
}
bool solve(int id, long double R) {
  t = 0;
  if (id <= n)
    st[++t] = node(-P, 1), st[++t] = node(P, -1);
  else {
    for (int i = 1; i <= n; i++) {
      if (dis(p[i], p[id]) > R + R - eps) continue;
      long double a = angle(p[id], p[i], R),
                  b = atan2(p[i].y - p[id].y, p[i].x - p[id].x);
      long double c = (b - a < -P ? b - a + 2 * P : b - a),
                  d = (a + b > P ? a + b - 2 * P : a + b);
      if (c < d)
        st[++t] = node(c, 1), st[++t] = node(d, -1);
      else
        st[++t] = node(-P, 1), st[++t] = node(d, -1), st[++t] = node(c, 1),
        st[++t] = node(P, -1);
    }
  }
  for (int i = n + 1; i <= n + m; i++) {
    if (dis(p[i], p[id]) > R + R - eps) continue;
    long double a = angle(p[id], p[i], R),
                b = atan2(p[i].y - p[id].y, p[i].x - p[id].x);
    long double c = (b - a < -P ? b - a + 2 * P : b - a),
                d = (a + b > P ? a + b - 2 * P : a + b);
    if (c < d)
      st[++t] = node(c, -10000), st[++t] = node(d, 10000);
    else
      st[++t] = node(-P, -10000), st[++t] = node(d, 10000),
      st[++t] = node(c, -10000), st[++t] = node(P, 10000);
  }
  sort(st + 1, st + t + 1);
  int tmp = 0;
  for (int i = 1; i <= t; i++) {
    if (tmp > 0 && i != 1 && st[i].w - st[i - 1].w > eps) return true;
    tmp += st[i].id;
  }
  return false;
}
int main() {
  scanf("%d %d", &n, &m);
  if (m == 1) return puts("-1"), 0;
  for (int i = 1, x, y; i <= n; i++)
    scanf("%d %d", &x, &y), p[i].x = x, p[i].y = y;
  for (int i = 1, x, y; i <= m; i++)
    scanf("%d %d", &x, &y), p[n + i].x = x, p[n + i].y = y;
  long double l = 0, r, mid;
  for (int i = 1; i <= n + m; i++)
    if (solve(i, l)) {
      r = 1e9;
      while (1e-5 < r - l) {
        mid = (l + r) / 2;
        if (solve(i, mid))
          l = mid;
        else
          r = mid;
      }
    }
  if (l > 1e9 - 1)
    puts("-1");
  else
    cout << fixed << setprecision(18) << l;
}
