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;
pair<long double, long double> operator+(
    const pair<long double, long double> &a,
    const pair<long double, long double> &b) {
  return make_pair(a.first + b.first, a.second + b.second);
}
pair<long double, long double> operator-(
    const pair<long double, long double> &a,
    const pair<long double, long double> &b) {
  return make_pair(a.first - b.first, a.second - b.second);
}
pair<long double, long double> operator*(
    const pair<long double, long double> &a, long double k) {
  return make_pair(a.first * k, a.second * k);
}
pair<long double, long double> operator/(
    const pair<long double, long double> &a, long double k) {
  return make_pair(a.first / k, a.second / k);
}
long double dot(const pair<long double, long double> &a,
                const pair<long double, long double> &b) {
  return a.first * b.first + a.second * b.second;
}
long double det(const pair<long double, long double> &a,
                const pair<long double, long double> &b) {
  return a.first * b.second - b.first * a.second;
}
long double len2(const pair<long double, long double> &a) {
  return a.first * a.first + a.second * a.second;
}
long double len(const pair<long double, long double> &a) {
  return sqrt(len2(a));
}
long double At(const pair<long double, long double> &a) {
  return atan2(a.second, a.first);
}
const long double EPS = 1E-10;
pair<long double, long double> intersect(
    const pair<long double, long double> &a,
    const pair<long double, long double> &b,
    const pair<long double, long double> &c,
    const pair<long double, long double> &d) {
  long double k1 = det(b - a, c - a), k2 = det(b - a, d - a);
  if (fabs(k1 - k2) < EPS) return make_pair(-1, -1);
  return (d * k1 - c * k2) / (k1 - k2);
}
const int MAXN = 1000 + 10;
struct HP {
  pair<long double, long double> first, v;
  long double agl;
  int id;
  HP() {}
  HP(pair<long double, long double> first, pair<long double, long double> v,
     int id)
      : first(first), v(v), id(id) {
    agl = atan2(v.second, v.first);
  }
  bool operator<(const HP &b) const { return agl < b.agl; }
  void Print() {
    cout << first.first << " " << first.second << " " << first.first + v.first
         << " " << first.second + v.second << endl;
  }
} h[MAXN];
map<pair<int, int>, int> t;
int n, N, W, H, tot;
pair<long double, long double> p[MAXN];
bool ml[MAXN];
void init() {
  scanf("%d%d%d", &W, &H, &N);
  for (int i = 1; i <= N; ++i) {
    pair<int, int> w;
    cin >> w.first >> w.second;
    ++t[w];
  }
  for (map<pair<int, int>, int>::iterator it = t.begin(); it != t.end(); ++it)
    p[++n] = it->first, ml[n] = it->second > 1;
}
long double Count(int l, int r, int w, int id) {
  if (ml[h[w].id] || !h[w].id) return 0;
  static int head, tail, que[MAXN];
  static pair<long double, long double> b[MAXN];
  head = 1, tail = 1, que[1] = l;
  for (int i = l + 1;; ++i) {
    if (i > tot) i = 1;
    if (i == w) continue;
    while (head < tail && det(h[i].v, b[tail - 1] - h[i].first) < EPS) --tail;
    while (head < tail && det(h[i].v, b[head] - h[i].first) < EPS) ++head;
    que[++tail] = i;
    if (head < tail && fabs(h[que[tail - 1]].agl - h[i].agl) < EPS) {
      --tail;
      if (det(h[que[tail]].v, h[i].first - h[que[tail]].first) > 0)
        que[tail] = i;
    }
    if (head < tail)
      b[tail - 1] = intersect(
          h[que[tail - 1]].first, h[que[tail - 1]].first + h[que[tail - 1]].v,
          h[que[tail]].first, h[que[tail]].first + h[que[tail]].v);
    if (i == r) break;
  }
  long double Ririchiyo = 0;
  for (int i = head; i < tail; ++i) {
    if (b[i] != pair<long double, long double>(-1, -1) && b[i].first >= 0 &&
        b[i].first <= (long double)W && b[i].second >= 0 &&
        b[i].second <= (long double)H)
      Ririchiyo = max(Ririchiyo, len(b[i] - p[id]));
  }
  return Ririchiyo;
}
long double Calc(int id) {
  tot = 0;
  h[++tot] = HP(pair<long double, long double>(0, 0),
                pair<long double, long double>(W, 0), 0);
  h[++tot] = HP(pair<long double, long double>(W, 0),
                pair<long double, long double>(0, H), 0);
  h[++tot] = HP(pair<long double, long double>(W, H),
                pair<long double, long double>(-W, 0), 0);
  h[++tot] = HP(pair<long double, long double>(0, H),
                pair<long double, long double>(0, -H), 0);
  for (int i = 1; i <= n; ++i)
    if (i != id) {
      pair<long double, long double> v = p[i] - p[id], o = p[id] + v * 0.5;
      swap(v.first, v.second);
      v.first = -v.first;
      h[++tot] = HP(o, v, i);
    }
  sort(h + 1, h + tot + 1);
  static int head, tail, que[MAXN];
  static pair<long double, long double> b[MAXN];
  head = tail = 1, que[1] = 1;
  for (int i = 2; i <= tot; ++i) {
    while (head < tail && det(h[i].v, b[tail - 1] - h[i].first) < EPS) --tail;
    while (head < tail && det(h[i].v, b[head] - h[i].first) < EPS) ++head;
    que[++tail] = i;
    if (head < tail && fabs(h[que[tail - 1]].agl - h[i].agl) < EPS) {
      --tail;
      if (det(h[que[tail]].v, h[i].first - h[que[tail]].first) > 0)
        que[tail] = i;
    }
    if (head < tail)
      b[tail - 1] = intersect(
          h[que[tail - 1]].first, h[que[tail - 1]].first + h[que[tail - 1]].v,
          h[que[tail]].first, h[que[tail]].first + h[que[tail]].v);
  }
  while (head < tail &&
         det(h[que[head]].v, b[tail - 1] - h[que[head]].first) < 0)
    --tail;
  b[tail] = intersect(h[que[head]].first, h[que[head]].first + h[que[head]].v,
                      h[que[tail]].first, h[que[tail]].first + h[que[tail]].v);
  long double Ririchiyo = 0;
  if (ml[id]) {
    for (int i = head; i <= tail; ++i)
      Ririchiyo = max(Ririchiyo, len(b[i] - p[id]));
  }
  int m = tail - head + 1;
  for (int i = 1, j = head; i <= m; ++i, ++j) que[i] = que[j];
  for (int i = 1; i + 2 <= m; ++i)
    Ririchiyo = max(Ririchiyo, Count(que[i], que[i + 2], que[i + 1], id));
  Ririchiyo = max(Ririchiyo, Count(que[m - 1], que[1], que[m], id));
  Ririchiyo = max(Ririchiyo, Count(que[m], que[2], que[1], id));
  return Ririchiyo;
}
void solve() {
  long double Tohka = 0;
  cout << setprecision(10);
  for (int i = 1; i <= n; ++i) Tohka = max(Tohka, Calc(i));
  cout << fixed << Tohka << endl;
}
int main() {
  init();
  solve();
  fclose(stdin);
  fclose(stdout);
  return 0;
}