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 int N = 2222;
const long long INF = 1e9 + 19;
bool equalll(long double a, long double b) { return abs(a - b) < 1e-9; }
bool lesss(long double a, long double b) { return a < b && !equalll(a, b); }
bool lessE(long double a, long double b) { return a < b || equalll(a, b); }
bool equal3(long double a, long double b) { return abs(a - b) < 1e-3; }
bool less3(long double a, long double b) { return a < b && !equal3(a, b); }
bool lessE3(long double a, long double b) { return a < b || equal3(a, b); }
struct pt {
  long double x, y;
  pt() {}
  pt(long double xx, long double yy) : x(xx), y(yy) {}
  pt operator+(pt A) { return pt(x + A.x, y + A.y); }
  pt operator-(pt A) { return pt(x - A.x, y - A.y); }
  long double operator*(pt A) { return x * A.y - y * A.x; }
  long double operator%(pt A) { return x * A.x + y * A.y; }
  pt operator*(long double k) { return pt(x * k, y * k); }
  pt operator/(long double k) {
    assert(!equalll(k, 0));
    return pt(x / k, y / k);
  }
  bool operator==(pt A) { return equalll(x, A.x) && equalll(y, A.y); }
  bool operator!=(pt A) { return !equalll(x, A.x) && !equalll(y, A.y); }
  long double len() { return sqrt(x * x + y * y); }
  long double len2() { return x * x + y * y; }
  pt norm() { return (*this) / len(); }
  pt rotate() { return pt(-y, x); }
  void read() {
    double xx, yy;
    scanf("%lf%lf", &xx, &yy);
    x = xx;
    y = yy;
  }
  void Epr() { cerr << "x y " << x << " " << y << endl; }
};
struct Line {
  pt O, v;
  Line() {}
  Line(pt A, pt B) : O(A), v((B - A).norm()) {}
  pt operator*(Line l) {
    pt u = l.v.rotate();
    double k = (l.O - O) % u / (v % u);
    return O + v * k;
  }
};
int n, m;
pt red[N];
pt blue[N];
vector<pt> convexHull(vector<pt> p) {
  for (int i = 0; i < (int)p.size(); i++) {
    if (lesss(p[i].y, p[0].y) ||
        (equalll(p[i].y, p[0].y) && lesss(p[i].x, p[0].x)))
      swap(p[i], p[0]);
  }
  pt O = p[0];
  sort(p.begin() + 1, p.end(), [&](pt A, pt B) {
    pt v = A - O;
    pt u = B - O;
    return lesss(0, v * u) || (equalll(v * u, 0) && lesss(v.len(), u.len()));
  });
  vector<pt> st;
  for (auto A : p) {
    for (; st.size() >= 2 &&
           lessE((st[(int)st.size() - 1] - st[(int)st.size() - 2]) *
                     (A - st.back()),
                 0);
         st.pop_back())
      ;
    st.push_back(A);
  }
  return st;
}
bool onLine(pt A, pt B, pt C) { return (equalll((B - A) * (C - A), 0)); }
bool onSeg(pt A, pt B, pt C) {
  return onLine(A, B, C) && (C - A) % (C - B) < 0;
}
int checkInside(vector<pt> &ch, pt A, int &id1, int &id2) {
  if (lesss(0, (A - ch[0]) * (ch[1] - ch[0]))) return 0;
  if (lesss(0, (ch.back() - ch[0]) * (A - ch[0]))) return 0;
  if (onSeg(ch[0], ch[1], A)) {
    id1 = 0;
    id2 = 1;
    return 1;
  }
  if (onSeg(ch.back(), ch[0], A)) {
    id1 = (int)ch.size() - 1;
    id2 = 0;
    return 1;
  }
  int l = 1;
  int r = (int)ch.size() - 1;
  while (r - l > 1) {
    int mid = (l + r) / 2;
    if ((A - ch[0]) * (ch[mid] - ch[0]) > 0)
      r = mid;
    else
      l = mid;
  }
  if (equalll(0, (ch[r] - ch[l]) * (A - ch[l]))) {
    assert(onSeg(ch[l], ch[r], A));
    id1 = l;
    id2 = r;
    return 1;
  }
  if (lesss(0, (ch[r] - ch[l]) * (A - ch[l]))) return 2;
  return 0;
}
long double distToLine(pt A, pt B, pt C) {
  pt v = (B - A).norm();
  return abs((C - A) * v);
}
bool cmpAng(pt dir, pt v, pt u) {
  bool partV = lesss(dir * v, 0) || (equalll(dir * v, 0) && lesss(dir % v, 0));
  bool partU = lesss(dir * u, 0) || (equalll(dir * u, 0) && lesss(dir % u, 0));
  if (partV != partU) return partV < partU;
  return lesss(0, v * u);
}
void read() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) {
    red[i].read();
  }
  for (int i = 0; i < m; i++) {
    blue[i].read();
  }
}
Line serPer(pt A, pt B) {
  pt C = (A + B) / 2;
  pt v = (B - A).rotate();
  return Line(C, C + v);
}
long double answer = INF;
const int T = 10000;
pt shift;
void updAnswer(pt A, pt B, pt C) {
  A = A + shift;
  B = B + shift;
  C = C + shift;
  long double cand = distToLine(A, B, C);
  answer = min(answer, cand);
  return;
  if (equalll(cand, 0)) {
    answer = min(answer, cand);
    return;
  }
  pt va = (A - C);
  pt vb = (B - C);
  va = va * T / va.len2();
  vb = vb * T / vb.len2();
  A = C + va;
  B = C + vb;
  Line ab = serPer(A, B);
  Line ac = serPer(A, C);
  pt O = ab * ac;
  long double R = (O - A).len();
  assert(equalll(R, (O - C).len()));
  assert(equalll(R, (O - B).len()));
  bool flagRed = 0;
  bool flagBlue = 1;
  for (int i = 0; i < m; i++) {
    flagBlue &= lessE3(R, (O - blue[i]).len());
  }
  for (int i = 0; i < n; i++) {
    flagRed |= lessE3((O - red[i]).len(), R);
  }
  assert(flagRed && flagBlue);
  answer = min(answer, cand);
}
double solve() {
  answer = INF;
  if (m <= 2) {
    return -1;
  }
  for (int i = 0; i < m; i++) {
    vector<pt> p;
    pt O = blue[i];
    shift = blue[i];
    for (int j = 0; j < m; j++) {
      if (i == j) {
        p.push_back(pt(0, 0));
        continue;
      }
      pt v = blue[j] - O;
      v = v / v.len2();
      p.push_back(v * T);
    }
    vector<pt> b;
    for (int j = 0; j < n; j++) {
      pt v = red[j] - O;
      b.push_back(v / v.len2() * T);
    }
    O = pt(0, 0);
    auto ch = convexHull(p);
    if ((int)ch.size() <= 2) {
      answer = 0;
      continue;
    }
    int g = 0;
    for (; (1ll << (g + 1)) < (int)ch.size(); g++)
      ;
    vector<int> cof((int)ch.size() + 1);
    for (auto A : b) {
      int id1, id2;
      int res = checkInside(ch, A, id1, id2);
      if (res == 1) {
        updAnswer(ch[id1], ch[id2], O);
      }
      if (res == 0) {
        int lt = 0;
        for (int j = g; j >= 0; j--) {
          int best = lt;
          for (int tt = -1; tt <= 1; tt += 2) {
            int np =
                ((lt + ((1ll << (j)) * tt)) + (int)ch.size()) % (int)ch.size();
            if ((ch[best] - A) * (ch[np] - A) > 0) {
              best = np;
            }
          }
          lt = best;
        }
        int rt = 0;
        for (int j = g; j >= 0; j--) {
          int best = rt;
          for (int tt = -1; tt <= 1; tt += 2) {
            int np =
                ((rt + ((1ll << (j)) * tt)) + (int)ch.size()) % (int)ch.size();
            if ((ch[best] - A) * (ch[np] - A) < 0) {
              best = np;
            }
          }
          rt = best;
        }
        updAnswer(ch[lt], A, O);
        updAnswer(ch[rt], A, O);
        assert(lt != rt);
        if (lt < rt) {
          cof[lt]++;
          cof[rt]--;
        } else {
          cof[lt]++;
          cof[(int)ch.size()]--;
          cof[0]++;
          cof[rt]--;
        }
      }
    }
    int bal = 0;
    for (int j = 0; j < (int)ch.size(); j++) {
      bal += cof[j];
      if (bal > 0) {
        updAnswer(ch[j], ch[(j + 1) % (int)ch.size()], O);
      }
    }
  }
  if (equalll(answer, 0)) {
    return -1;
  } else {
    return T * 1.0 / answer / 2;
  }
}
void genTest() {
  set<pair<int, int>> q;
  int C = 10;
  n = rand() % C + 1;
  m = rand() % C + 1;
  int BB = 10000;
  for (int i = 0; i < n; i++) {
    int x, y;
    for (;;) {
      x = rand() % BB;
      y = rand() % BB;
      if (q.count(make_pair(x, y)) == 0) {
        q.insert(make_pair(x, y));
        break;
      }
    }
    red[i] = pt(x, y);
  }
  for (int i = 0; i < m; i++) {
    int x, y;
    for (;;) {
      x = rand() % BB;
      y = rand() % BB;
      if (q.count(make_pair(x, y)) == 0) {
        q.insert(make_pair(x, y));
        break;
      }
    }
    blue[i] = pt(x, y);
  }
}
void printTest() {
  cout << n << " " << m << endl;
  for (int i = 0; i < n; i++) cerr << red[i].x << " " << red[i].y << endl;
  for (int i = 0; i < m; i++) cerr << blue[i].x << " " << blue[i].y << endl;
}
void stress() {
  for (int tt = 0;; tt++) {
    cerr << "tt"
         << " = " << tt << endl;
    genTest();
    printTest();
    solve();
  }
}
int main() {
  if (1) {
    read();
    printf("%.17f\n", solve());
  } else {
    stress();
  }
  return 0;
}
