#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) f = ch == '-' ? -1 : f, ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x * f;
}
const int N = 1005;
const int V = N << 1;
const long double EPS = 1e-4;
const long double INF = 1e+9;
const long double pi = acosl(-1);
const int ERROR = -100000;
inline bool equ(long double x, long double y) { return fabs(x - y) <= EPS; }
inline int sgn(long double x) { return equ(x, 0.) ? 0 : (x < 0 ? -1 : 1); }
struct P {
  long double x, y;
  inline P(long double x_ = 0., long double y_ = 0.) { x = x_, y = y_; }
  inline P operator+(P const p) const { return P(x + p.x, y + p.y); }
  inline P operator-(P const p) const { return P(x - p.x, y - p.y); }
  inline P operator*(long double const k) const { return P(x * k, y * k); }
  inline long double operator*(P const p) const { return x * p.x + y * p.y; }
  inline long double operator^(P const p) const { return x * p.y - y * p.x; }
  inline long double mod2() const { return (*this) * (*this); }
  inline long double mod() const { return sqrt(fabs(mod2())); }
  inline long double arg() { return atan2(y, x); }
} pts[V];
int n, m;
long double ans;
pair<long double, int> evt[V << 1];
long double dist[V][V];
inline bool cmp(pair<long double, int> p, pair<long double, int> q) {
  return p.first < q.first;
}
inline void adjust(long double &x) {
  x = x > pi ? x - pi * 2 : x < -pi ? x + pi * 2 : x;
}
inline bool judge(int id, long double r) {
  int cur = id <= n, cnt = 0, ret = 0;
  for (int i = 1; i <= n + m; ++i)
    if (i != id) {
      long double d = dist[i][id];
      if (sgn(d - 2. * r) > 0) continue;
      long double alpha = acosl(d / (2. * r)) - 1e-30 * (i > n),
                  mid = (pts[i] - pts[id]).arg();
      long double st = mid - alpha, en = mid + alpha;
      adjust(st), adjust(en);
      int cst = i <= n ? 1 : ERROR;
      cur += (sgn(en - st) < 0) * cst, evt[++cnt] = make_pair(st, cst),
                                       evt[++cnt] = make_pair(en, -cst);
    }
  ret = cur, sort(evt + 1, evt + 1 + cnt, cmp);
  for (int i = 1; i <= cnt; ++i) ret = max(ret, cur += evt[i].second);
  return ret >= 1;
}
inline void binary_search(int id) {
  long double l = ans, r = ans;
  for (; judge(id, r) && r <= INF; r *= 2)
    ;
  for (long double mid; sgn(r - l) >= 0;) {
    mid = (l + r) * .5;
    if (judge(id, mid))
      l = (ans = mid) + EPS;
    else
      r = mid - EPS;
  }
}
int main() {
  n = read(), m = read();
  for (int i = 1, x, y; i <= n + m; ++i)
    x = read(), y = read(), pts[i] = P(x, y);
  random_shuffle(pts + 1, pts + 1 + n),
      random_shuffle(pts + 1 + n, pts + 1 + n + m);
  for (int i = 1; i <= n + m; ++i)
    for (int j = 1; j <= n + m; ++j) dist[i][j] = (pts[i] - pts[j]).mod();
  ans = 1.;
  for (int i = 1; i <= n; ++i) {
    if (!judge(i, ans)) continue;
    binary_search(i);
  }
  for (int i = n + 1; i <= n + m; ++i) {
    if (!judge(i, ans)) continue;
    binary_search(i);
  }
  if (ans < INF * .5)
    printf("%.10lf\n", (double)ans);
  else
    printf("-1\n");
  return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) f = ch == '-' ? -1 : f, ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x * f;
}
const int N = 1005;
const int V = N << 1;
const long double EPS = 1e-4;
const long double INF = 1e+9;
const long double pi = acosl(-1);
const int ERROR = -100000;
inline bool equ(long double x, long double y) { return fabs(x - y) <= EPS; }
inline int sgn(long double x) { return equ(x, 0.) ? 0 : (x < 0 ? -1 : 1); }
struct P {
  long double x, y;
  inline P(long double x_ = 0., long double y_ = 0.) { x = x_, y = y_; }
  inline P operator+(P const p) const { return P(x + p.x, y + p.y); }
  inline P operator-(P const p) const { return P(x - p.x, y - p.y); }
  inline P operator*(long double const k) const { return P(x * k, y * k); }
  inline long double operator*(P const p) const { return x * p.x + y * p.y; }
  inline long double operator^(P const p) const { return x * p.y - y * p.x; }
  inline long double mod2() const { return (*this) * (*this); }
  inline long double mod() const { return sqrt(fabs(mod2())); }
  inline long double arg() { return atan2(y, x); }
} pts[V];
int n, m;
long double ans;
pair<long double, int> evt[V << 1];
long double dist[V][V];
inline bool cmp(pair<long double, int> p, pair<long double, int> q) {
  return p.first < q.first;
}
inline void adjust(long double &x) {
  x = x > pi ? x - pi * 2 : x < -pi ? x + pi * 2 : x;
}
inline bool judge(int id, long double r) {
  int cur = id <= n, cnt = 0, ret = 0;
  for (int i = 1; i <= n + m; ++i)
    if (i != id) {
      long double d = dist[i][id];
      if (sgn(d - 2. * r) > 0) continue;
      long double alpha = acosl(d / (2. * r)) - 1e-30 * (i > n),
                  mid = (pts[i] - pts[id]).arg();
      long double st = mid - alpha, en = mid + alpha;
      adjust(st), adjust(en);
      int cst = i <= n ? 1 : ERROR;
      cur += (sgn(en - st) < 0) * cst, evt[++cnt] = make_pair(st, cst),
                                       evt[++cnt] = make_pair(en, -cst);
    }
  ret = cur, sort(evt + 1, evt + 1 + cnt, cmp);
  for (int i = 1; i <= cnt; ++i) ret = max(ret, cur += evt[i].second);
  return ret >= 1;
}
inline void binary_search(int id) {
  long double l = ans, r = ans;
  for (; judge(id, r) && r <= INF; r *= 2)
    ;
  for (long double mid; sgn(r - l) >= 0;) {
    mid = (l + r) * .5;
    if (judge(id, mid))
      l = (ans = mid) + EPS;
    else
      r = mid - EPS;
  }
}
int main() {
  n = read(), m = read();
  for (int i = 1, x, y; i <= n + m; ++i)
    x = read(), y = read(), pts[i] = P(x, y);
  random_shuffle(pts + 1, pts + 1 + n),
      random_shuffle(pts + 1 + n, pts + 1 + n + m);
  for (int i = 1; i <= n + m; ++i)
    for (int j = 1; j <= n + m; ++j) dist[i][j] = (pts[i] - pts[j]).mod();
  ans = 1.;
  for (int i = 1; i <= n; ++i) {
    if (!judge(i, ans)) continue;
    binary_search(i);
  }
  for (int i = n + 1; i <= n + m; ++i) {
    if (!judge(i, ans)) continue;
    binary_search(i);
  }
  if (ans < INF * .5)
    printf("%.10lf\n", (double)ans);
  else
    printf("-1\n");
  return 0;
}

```