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;
const int INF = 0x3fffffff;
const int SINF = 0x7fffffff;
const long long LINF = 0x3fffffffffffffff;
const long long SLINF = 0x7fffffffffffffff;
const int MAXN = 1007;
const int MAXT = 50;
const int MAX = 2000007;
const double ANS_EPS = 3e-9;
const double EPS = 1e-13;
const double PI = acos(-1.0);
struct cT {
  cT() : x(0), y(0) {}
  cT(int _x, int _y) : x(_x), y(_y) {}
  int x, y;
} cr[MAXN];
struct eT {
  void setd(double _d, int _x) { deg = _d, x = _x; }
  bool operator<(const eT &a) const { return deg < a.deg; }
  double deg;
  int x;
} ev[MAXN * 5 + 107];
int W, H, n;
int ke;
int p[MAXN];
void init();
void input();
void work();
bool check(int i, double tst);
void add(double d1, double d2, double od, int x);
double getdist(const cT &a, const cT &b) {
  return sqrt(
      static_cast<double>(static_cast<long long>(a.x - b.x) * (a.x - b.x) +
                          static_cast<long long>(a.y - b.y) * (a.y - b.y)));
}
int main() {
  init();
  input();
  work();
}
void init() { ios::sync_with_stdio(false); }
void input() {
  scanf("%d%d%d", &W, &H, &n);
  for (int i = 1; i <= n; ++i) scanf("%d%d", &cr[i].x, &cr[i].y);
}
void work() {
  double ans = 0;
  double l, r, mid;
  for (int i = 1; i <= n; ++i) p[i] = i;
  random_shuffle(p + 1, p + 1 + n);
  for (int _i = 1; _i <= n; ++_i) {
    int i = p[_i];
    if (!check(i, ans + ANS_EPS)) continue;
    l = ans, r = MAX;
    for (int _t = 0; _t < MAXT; ++_t) {
      mid = l + (r - l) / 2;
      if (check(i, mid))
        l = mid;
      else
        r = mid;
    }
    ans = l;
  }
  cout << setprecision(15) << fixed << ans << endl;
}
bool check(int i, double tst) {
  ke = 0;
  add(tst, W - cr[i].x, 0, MAXN);
  add(tst, H - cr[i].y, PI / 2, MAXN);
  add(tst, cr[i].x, PI, MAXN);
  add(tst, cr[i].y, PI * 3 / 2, MAXN);
  for (int j = 1; j <= n; ++j) {
    if (i ^ j) {
      if (cr[i].x == cr[j].x && cr[i].y == cr[j].y) {
      } else
        add(tst, getdist(cr[i], cr[j]) / 2,
            atan2(cr[j].y - cr[i].y, cr[j].x - cr[i].x), 1);
    }
  }
  sort(ev + 1, ev + 1 + ke);
  double pre = 0;
  int nc = 0;
  for (int j = 1; j <= ke; ++j) {
    if (nc <= 1 && ev[j].deg - pre > EPS) return true;
    nc += ev[j].x;
    pre = ev[j].deg;
  }
  if (2 * PI - pre > EPS) return true;
  return false;
}
void add(double d1, double d2, double od, int x) {
  if (d1 - d2 < EPS) return;
  double dd = acos(d2 / d1);
  double dl = od - dd, dr = od + dd;
  while (dl < -EPS) dl += 2 * PI;
  while (dl > 2 * PI - EPS) dl -= 2 * PI;
  while (dr < -EPS) dr += 2 * PI;
  while (dr > 2 * PI - EPS) dr -= 2 * PI;
  if (dr - dl > -EPS) {
    ev[++ke].setd(dl, x);
    ev[++ke].setd(dr, -x);
  } else {
    ev[++ke].setd(0, x);
    ev[++ke].setd(dr, -x);
    ev[++ke].setd(dl, x);
    ev[++ke].setd(2 * PI, -x);
  }
}