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 power(int a, int b, int c, int ans = 1) {
  for (; b; b >>= 1, a = (long long)a * a % c)
    if (b & 1) ans = (long long)ans * a % c;
  return ans;
}
template <class T>
inline void smax(T &first, const T &second) {
  if (first < second) first = second;
}
using pdd = pair<double, double>;
pdd p[101010], ori[101010], q[101010];
int nn, first[101010], second[101010];
double dist(const int &first, const int &second, const double &x1,
            const double &y1) {
  return sqrt((first - x1) * (first - x1) + (second - y1) * (second - y1));
}
const double eps = 1e-10;
int sgn(const double &x1, const double &y1, const double &x2,
        const double &y2) {
  double s = x1 * y2 - y1 * x2;
  return (s > eps) - (s < -eps);
}
pdd calc(const pdd &a, const pdd &b, const pdd &c, const pdd &d) {
  double t = ((b.first - a.first) * (c.second - a.second) -
              (b.second - a.second) * (c.first - a.first)) /
             ((b.first - a.first) * d.second - (b.second - a.second) * d.first);
  return make_pair(c.first - t * d.first, c.second - t * d.second);
}
void cut(double x1, double y1, double x2, double y2) {
  if (!nn) return;
  int m = 0, pre = sgn(x2, y2, p[0].first - x1, p[0].second - y1);
  p[nn] = p[0];
  for (int i = 0; i < nn; i++) {
    int now = sgn(x2, y2, p[i + 1].first - x1, p[i + 1].second - y1);
    if (!now)
      q[m++] = p[i + 1];
    else if (!pre) {
      if (now > 0) q[m++] = p[i + 1];
    } else if (pre > 0) {
      if (now < 0)
        q[m++] = calc(p[i], p[i + 1], make_pair(x1, y1), make_pair(x2, y2));
      else
        q[m++] = p[i + 1];
    } else {
      if (now > 0) {
        q[m++] = calc(p[i], p[i + 1], make_pair(x1, y1), make_pair(x2, y2));
        q[m++] = p[i + 1];
      }
    }
    pre = now;
  }
  nn = m;
  for (int i = 0; i < nn; i++) p[i] = q[i];
}
int main() {
  int w, h, n;
  cin >> w >> h >> n;
  for (int i = 0; i < n; i++) scanf("%d%d", first + i, second + i);
  double ans = 0;
  for (int i = 0; i < n; i++) {
    nn = 0;
    p[nn++] = make_pair(0, 0);
    p[nn++] = make_pair(w, 0);
    p[nn++] = make_pair(w, h);
    p[nn++] = make_pair(0, h);
    int ok = 0;
    for (int j = 0; j < n; j++)
      if (i != j) {
        if (first[i] == first[j] && second[i] == second[j]) {
          ok = 1;
          continue;
        }
        cut((first[i] + first[j]) * 0.5, (second[i] + second[j]) * 0.5,
            -(second[j] - second[i]), first[j] - first[i]);
      }
    if (ok) {
      for (int j = 0; j < nn; j++)
        smax(ans, dist(first[i], second[i], p[j].first, p[j].second));
      continue;
    }
    vector<int> id;
    p[nn] = p[0];
    for (int j = 0; j < nn; j++) {
      int ii = -1;
      double cur = 1e100, tmp;
      double xx = (p[j].first + p[j + 1].first) / 2,
             yy = (p[j].second + p[j + 1].second) / 2;
      for (int k = 0; k < n; k++)
        if (k != i) {
          if (first[k] == first[i] && second[k] == second[i]) continue;
          if ((tmp = dist(first[k], second[k], xx, yy)) < cur)
            cur = tmp, ii = k;
        }
      id.emplace_back(ii);
    }
    sort(id.begin(), id.end());
    id.resize(unique(id.begin(), id.end()) - id.begin());
    int m = nn;
    for (int j = 0; j < nn; j++) ori[j] = p[j];
    for (int j : id) {
      nn = m;
      for (int k = 0; k < m; k++) p[k] = ori[k];
      for (int k : id)
        if (j != k) {
          cut((first[j] + first[k]) * 0.5, (second[j] + second[k]) * 0.5,
              -(second[k] - second[j]), first[k] - first[j]);
        }
      for (int k = 0; k < nn; k++)
        smax(ans, dist(first[j], second[j], p[k].first, p[k].second));
    }
  }
  cout << fixed << setprecision(20) << ans << endl;
  return 0;
}