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:512000000")
using namespace std;
template <class T>
inline T sqr(T x) {
  return x * x;
}
template <class T>
inline string tostr(const T& x) {
  stringstream ss;
  ss << x;
  return ss.str();
}
inline long long parse(const string& s) {
  stringstream ss(s);
  long long x;
  ss >> x;
  return x;
}
const double EPS = 1e-9;
const int INF = 1000 * 1000 * 1000;
const char CINF = 102;
const long long LINF = INF * 1ll * INF;
const double DINF = 1e200;
const double PI = 3.1415926535897932384626433832795l;
int gcd(int a, int b) { return a ? gcd(b % a, a) : b; }
long long gcd(long long a, long long b) { return a ? gcd(b % a, a) : b; }
long long powmod(long long a, long long p, long long m) {
  long long r = 1;
  while (p) {
    if (p & 1) r = r * a % m;
    p >>= 1;
    a = a * a % m;
  }
  return r;
}
double W, H;
vector<pair<double, double> > A;
double P[1100][1100];
double Q[1100][1100];
bool check(double R, int i0) {
  vector<pair<double, int> > evs;
  int D = 0;
  pair<double, double> v = A[i0];
  if (v.first < R) {
    double a = PI - acos(v.first / R);
    D += 2;
    evs.push_back(make_pair(-a, -2));
    evs.push_back(make_pair(+a, +2));
  }
  if (W - v.first < R) {
    double a = acos((W - v.first) / R);
    evs.push_back(make_pair(-a, +2));
    evs.push_back(make_pair(+a, -2));
  }
  if (v.second < R) {
    double a = acos(v.second / R);
    evs.push_back(make_pair(-PI / 2 - a, +2));
    evs.push_back(make_pair(-PI / 2 + a, -2));
  }
  if (H - v.second < R) {
    double a = acos((H - v.second) / R);
    evs.push_back(make_pair(PI / 2 - a, +2));
    evs.push_back(make_pair(PI / 2 + a, -2));
  }
  for (int i = 0; i < (((int)(A).size())); ++i) {
    if (i == i0 || A[i] == v) continue;
    double d = P[i0][i];
    if (d >= R * 2) continue;
    double a = acos(d / 2 / R);
    double x = Q[i0][i];
    if (x + a > PI) x -= 2 * PI;
    if (x - a < -PI) {
      ++D;
      evs.push_back(make_pair(x + a, -1));
      evs.push_back(make_pair(2 * PI + x - a, +1));
    } else {
      evs.push_back(make_pair(x - a, +1));
      evs.push_back(make_pair(x + a, -1));
    }
  }
  sort((evs).begin(), (evs).end());
  if (D <= 1) return true;
  for (int i = 0; i < (((int)(evs).size())); ++i) {
    D += evs[i].second;
    if (D <= 1) return true;
  }
  return false;
}
int main() {
  int n;
  cin >> W >> H >> n;
  A.resize(n);
  for (int i = 0; i < (n); ++i) {
    cin >> A[i].first >> A[i].second;
  }
  random_shuffle((A).begin(), (A).end());
  for (int i = 0; i < (n); ++i) {
    for (int j = 0; j < (n); ++j) {
      if (i <= j)
        P[i][j] = P[j][i] =
            sqrt(sqr(A[i].first - A[j].first) + sqr(A[i].second - A[j].second));
      if (A[i] != A[j])
        Q[i][j] = atan2(A[j].second - A[i].second, A[j].first - A[i].first);
    }
  }
  double ans = 0;
  for (int i0 = 0; i0 < (((int)(A).size())); ++i0) {
    double a = ans, b = sqrt(sqr(H) + sqr(W));
    if (!check(a + EPS, i0)) continue;
    for (int qqq = 0; qqq < (60); ++qqq) {
      double r = (a + b) / 2;
      if (check(r, i0))
        a = r;
      else
        b = r;
    }
    ans = a;
  }
  printf("%.15lf", (double)ans);
  return 0;
}