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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target("avx")
#pragma GCC target("sse,sse2,sse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
long long read() {
  long long x = 0;
  bool q = 0;
  char c = getchar();
  while (!isdigit(c)) q |= (c == '-'), c = getchar();
  while (isdigit(c)) x = (x << 1) + (x << 3) + c - '0', c = getchar();
  return q ? -x : x;
}
void print(long long x, char q = '\n') {
  if (x < 0) putchar('-'), x = -x;
  if (x == 0) putchar('0');
  stack<char> s;
  s.push(q);
  while (x > 0) s.push(x % 10 + '0'), x /= 10;
  while (!s.empty()) putchar(s.top()), s.pop();
}
string read_s() {
  string s;
  char c = getchar();
  while (c == ' ' || c == '\t' || c == '\n') c = getchar();
  while (c != ' ' && c != '\t' && c != '\n') s += c, c = getchar();
  return s;
}
void print_s(string s, char q = '\n') {
  for (char c : s) putchar(c);
  putchar(q);
}
char read_c() {
  char c = getchar();
  while (c == ' ' || c == '\t' || c == '\n') c = getchar();
  return c;
}
const long double eps = 1e-12;
bool equal(long double x, long double y) { return x - eps < y && y < x + eps; }
struct Point {
  long double x, y;
  Point() {}
  Point(long double x, long double y) : x(x), y(y) {}
  Point operator+(const Point &t) const { return Point(x + t.x, y + t.y); }
  Point operator-(const Point &t) const { return Point(x - t.x, y - t.y); }
  Point operator*(const long double &t) { return Point(x * t, y * t); }
  long double operator*(const Point &t) const { return x * t.x + y * t.y; }
  Point operator/(const long double &t) const { return Point(x / t, y / t); }
  bool operator<(const Point &t) const {
    return fabs(x - t.x) < eps ? y < t.y : x < t.x;
  }
  bool operator==(const Point &t) const {
    return equal(x, t.x) && equal(y, t.y);
  }
  long double dist2() const { return x * x + y * y; }
  long double dist() const { return sqrt(dist2()); }
  Point rot90() const { return Point(-y, x); }
  Point rot(double t) const {
    return Point(x * cos(t) - y * sin(t), x * sin(t) + y * cos(t));
  }
  Point ProyInLine(const Point &a, const Point &b) const {
    return a + (b - a) * (((*this) - a) * (b - a)) / (b - a).dist2();
  }
  Point ProyInSeg(const Point &a, const Point &b) const {
    if ((b - a).dist2() < eps) return a;
    long double r = (((*this) - a) * (b - a)) / (b - a).dist2();
    if (r < 0) return a;
    if (r > 1) return b;
    return a + (b - a) * r;
  }
};
long double X, Y;
struct Bisector {
  long double A, B, C, ang, hyp;
  int id;
  Bisector(long double A, long double B, long double C, long double hyp)
      : A(A), B(B), C(C), hyp(hyp) {
    ang = atan2(A, B);
    id = -1;
  }
  Bisector(Point a, Point b, int id) : id(id) {
    Point mid = (a + b) * 0.5;
    A = b.x - a.x;
    B = b.y - a.y;
    C = -A * mid.x - B * mid.y;
    ang = atan2(A, B);
    hyp = hypot(A, B);
  }
  bool operator^(Bisector L) {
    long double det = A * L.B - L.A * B;
    X = -(C * L.B - B * L.C) / det, Y = (C * L.A - A * L.C) / det;
    return !equal(det, 0);
  }
};
bool operator<(const Bisector &a, const Bisector &b) {
  return equal(a.ang, b.ang) ? (b.hyp + eps < a.hyp) : (a.ang + eps < b.ang);
}
bool bad(Bisector a, Bisector b, Bisector c) {
  a ^ b;
  return c.A * X + c.B * Y + c.C > eps;
}
deque<Bisector> ConvexHull(vector<Bisector> bs) {
  sort(bs.begin(), bs.end());
  deque<Bisector> stk;
  for (auto u : bs) {
    while (!stk.empty() && equal(u.ang, stk.back().ang)) stk.pop_back();
    while (stk.size() > 1 && bad(stk[stk.size() - 2], stk.back(), u))
      stk.pop_back();
    stk.push_back(u);
  }
  bool q = 1;
  while (q) {
    q = 0;
    while (stk.size() > 2 && bad(stk[1], stk[0], stk.back()))
      stk.pop_front(), q = 1;
    while (stk.size() > 2 && bad(stk.back(), stk[stk.size() - 2], stk[0]))
      stk.pop_back(), q = 1;
  }
  return stk;
}
int W, H;
vector<Bisector> build(vector<Point> &p, int x) {
  vector<Bisector> ret;
  for (int i = 0; i < p.size(); i++)
    if (i != x) ret.push_back(Bisector(p[x], p[i], i));
  ret.push_back(Bisector(-1, 0, 0, 2 * p[x].x));
  ret.push_back(Bisector(0, -1, 0, 2 * p[x].y));
  ret.push_back(Bisector(1, 0, -W, 2 * W - 2 * p[x].x));
  ret.push_back(Bisector(0, 1, -H, 2 * H - 2 * p[x].y));
  return ret;
}
void Voronoi(vector<Point> &p, vector<Point> &ans) {
  int n = p.size();
  for (int i = 0; i < n; i++) {
    deque<Bisector> v = ConvexHull(build(p, i));
    for (int j = 1; j < v.size(); j++)
      if (v[j - 1] ^ v[j]) ans.push_back(Point(X, Y));
    if (v.back() ^ v[0]) ans.push_back(Point(X, Y));
  }
}
int main() {
  W = read(), H = read();
  int n = read();
  vector<Point> p(n), candi;
  for (int i = 0; i < n; i++) p[i].x = read(), p[i].y = read();
  for (int i = 0; i < n; i++) {
    deque<Bisector> v = ConvexHull(build(p, i));
    vector<Point> q;
    for (auto u : v)
      if (u.id != -1) q.push_back(p[u.id]);
    Voronoi(q, candi);
  }
  long double ans = 0;
  sort(candi.begin(), candi.end());
  candi.resize(unique(candi.begin(), candi.end()) - candi.begin());
  for (auto u : candi) {
    long double f = 1e18, g = 1e18;
    for (auto v : p) {
      long double x = (u - v).dist();
      if (x < f)
        g = f, f = x;
      else if (x < g)
        g = x;
    }
    ans = max(ans, g);
  }
  printf("%.17f\n", (double)ans);
}