Problem: Gena doesn't like geometry, so he asks you to solve this problem for him.

A rectangle with sides parallel to coordinate axes contains n dots. Let's consider some point of the plane. Let's count the distances from this point to the given n points. Let's sort these numbers in the non-decreasing order. We'll call the beauty of the point the second element of this array. If there are two mimimum elements in this array, the beaty will be equal to this minimum.

Find the maximum beauty of a point inside the given rectangle.

Input

The first line contains three integers w, h, n (1 ≤ w, h ≤ 106, 2 ≤ n ≤ 1000) — the lengths of the rectangle sides and the number of points. Next n lines contain two integers xi, yi (0 ≤ xi ≤ w, 0 ≤ yi ≤ h) each — the coordinates of a point. It is possible that it will be coincident points.

Output

Print a single number — the maximum beauty of a point with the absolute or relative error of at most 10 - 9.

Examples

Input

5 5 4
0 0
5 0
0 5
5 5


Output

4.99999999941792340


Input

5 5 3
4 0
2 5
4 1


Output

5.65685424744772010

Note

The point which beauty we need to find must have coordinates (x, y), where 0 ≤ x ≤ w, 0 ≤ y ≤ h. Some of the n points can coincide.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int dcmp(double p) {
  if (abs(p) < 1e-10) return 0;
  return p > 1e-10 ? 1 : -1;
}
struct Point {
  double x, y;
  Point() {}
  Point(double x, double y) : x(x), y(y) {}
  friend Point operator+(Point a, Point b) {
    return Point(a.x + b.x, a.y + b.y);
  }
  friend Point operator-(Point a, Point b) {
    return Point(a.x - b.x, a.y - b.y);
  }
  friend Point operator*(Point a, double p) { return Point(a.x * p, a.y * p); }
  friend Point operator/(Point a, double p) { return Point(a.x / p, a.y / p); }
  friend bool operator==(Point a, Point b) {
    return !dcmp(a.x - b.x) && !dcmp(a.y - b.y);
  }
  friend bool operator!=(Point a, Point b) {
    return dcmp(a.x - b.x) || dcmp(a.y - b.y);
  }
  friend bool operator<(const Point &a, const Point &b) {
    return dcmp(a.x - b.x) < 0 || (!dcmp(a.x - b.x) && dcmp(a.y - b.y) < 0);
  }
  void Read() { cin >> x >> y; }
} p[1010];
set<Point> Hash;
double Cross(Point a, Point b) { return a.x * b.y - a.y * b.x; }
double Dot(Point a, Point b) { return a.x * b.x + a.y * b.y; }
double Len(Point a) { return sqrt(Dot(a, a)); }
struct Line {
  Point p;
  Point v;
  double rad;
  Line() {}
  Line(Point p, Point v) : p(p), v(v) { rad = atan2(v.y, v.x); }
  friend bool operator<(const Line &a, const Line &b) { return a.rad < b.rad; }
} L[1010];
double W, H, ans;
int n;
Point getinter(Line a, Line b) {
  Point v = a.p - b.p;
  double t = Cross(b.v, v) / Cross(a.v, b.v);
  return a.p + a.v * t;
}
bool Left(Point p, Line L) { return dcmp(Cross(L.v, p - L.p)) > 0; }
void read() {
  cin >> W >> H >> n;
  for (int i = 1; i <= n; ++i) p[i].Read(), Hash.insert(p[i]);
}
double work(Point P) {
  double min1 = 1e100, min2 = 1e100;
  for (int i = 1; i <= n; ++i) {
    double len = Dot(P - p[i], P - p[i]);
    if (dcmp(min1 - len) >= 0)
      min2 = min1, min1 = len;
    else if (dcmp(min2 - len) >= 0)
      min2 = len;
  }
  return sqrt(min2);
}
void halfplane(Line *L, int n, Point *a, int &m, int type) {
  static Line queue[1010];
  static Point inter[1010];
  sort(L + 1, L + n + 1);
  int front = 1, rear = 1;
  queue[1] = L[1];
  for (int i = 2; i <= n; ++i) {
    while (front < rear && !Left(inter[rear - 1], L[i])) --rear;
    while (front < rear && !Left(inter[front], L[i])) ++front;
    queue[++rear] = L[i];
    if (!dcmp(Cross(L[i].v, queue[rear - 1].v)) && Left(L[i].p, queue[--rear]))
      queue[rear] = L[i];
    if (front < rear) inter[rear - 1] = getinter(queue[rear - 1], queue[rear]);
  }
  while (rear > front && !Left(inter[rear - 1], queue[front])) --rear;
  inter[rear] = getinter(queue[front], queue[rear]);
  for (int i = front; i <= rear; ++i) {
    ans = max(ans, work(inter[i]));
    if (!type) a[++m] = queue[i].p;
  }
}
void work() {
  Point A(W, H), B(0, H), C(0, 0), D(W, 0);
  static Point tmp[1010];
  for (int i = 1; i <= n; ++i) {
    int m = 0;
    L[++m] = Line(A, Point(-1, 0));
    L[++m] = Line(B, Point(0, -1));
    L[++m] = Line(C, Point(1, 0));
    L[++m] = Line(D, Point(0, 1));
    for (int j = 1; j <= n; ++j)
      if (p[i] != p[j]) {
        Point P = (p[i] + p[j]) / 2;
        Point v = p[j] - p[i];
        L[++m] = Line(P, Point(-v.y, v.x));
      }
    int cnt = 0;
    halfplane(L, m, tmp, cnt, 0);
    int tot = 0;
    for (int j = 1; j <= cnt; ++j) {
      tmp[j] = tmp[j] * 2 - p[i];
      if (Hash.count(tmp[j])) tmp[++tot] = tmp[j];
    }
    cnt = tot;
    for (int j = 1; j <= cnt; ++j) {
      m = 0;
      L[++m] = Line(A, Point(-1, 0));
      L[++m] = Line(B, Point(0, -1));
      L[++m] = Line(C, Point(1, 0));
      L[++m] = Line(D, Point(0, 1));
      for (int k = 1; k <= cnt; ++k)
        if (j != k) {
          Point P = (tmp[j] + tmp[k]) / 2;
          Point v = tmp[k] - tmp[j];
          L[++m] = Line(P, Point(-v.y, v.x));
        }
      int ncnt = 0;
      halfplane(L, m, tmp, ncnt, 1);
    }
  }
}
int main() {
  read();
  work();
  printf("%.20f\n", ans);
  return 0;
}