Problem: Many schoolchildren look for a job for the summer, and one day, when Gerald was still a schoolboy, he also decided to work in the summer. But as Gerald was quite an unusual schoolboy, he found quite unusual work. A certain Company agreed to pay him a certain sum of money if he draws them three identical circles on a plane. The circles must not interfere with each other (but they may touch each other). He can choose the centers of the circles only from the n options granted by the Company. He is free to choose the radius of the circles himself (all three radiuses must be equal), but please note that the larger the radius is, the more he gets paid. 

Help Gerald earn as much as possible.

Input

The first line contains a single integer n — the number of centers (3 ≤ n ≤ 3000). The following n lines each contain two integers xi, yi ( - 104 ≤ xi, yi ≤ 104) — the coordinates of potential circle centers, provided by the Company.

All given points are distinct.

Output

Print a single real number — maximum possible radius of circles. The answer will be accepted if its relative or absolute error doesn't exceed 10 - 6.

Examples

Input

3
0 1
1 0
1 1


Output

0.50000000000000000000


Input

7
2 -3
-2 -3
3 0
-3 -1
1 -2
2 -2
-1 0


Output

1.58113883008418980000

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int N, otoczka[3010];
pair<int, int> punkty[3010];
bool rozw[3010], flaga;
bool czy_wypukly(pair<int, int> a, pair<int, int> b, pair<int, int> c,
                 pair<int, int> d) {
  int a_x = a.first - b.first;
  int a_y = a.second - b.second;
  int b_x = c.first - d.first;
  int b_y = c.second - d.second;
  if (a_x * b_y - a_y * b_x > 0) {
    return 1;
  }
  return 0;
}
void wczytaj_punkty() {
  int x, y;
  scanf("%d", &N);
  for (int i = 0; i < N; ++i) {
    scanf("%d%d", &x, &y);
    punkty[i] = make_pair(x, y);
  }
}
int znajdz_otoczke() {
  int ile_punktow = 0;
  for (int i = 0; i < N; ++i) {
    if (rozw[i]) {
      ++ile_punktow;
    }
  }
  if (ile_punktow < 2) {
    return ile_punktow;
  }
  int i = 0;
  while (!rozw[i]) {
    ++i;
  }
  otoczka[0] = i;
  ++i;
  while (!rozw[i]) {
    ++i;
  }
  otoczka[1] = i;
  ++i;
  int rozmiar1 = 2;
  for (; i < N; ++i) {
    if (!rozw[i]) {
      continue;
    }
    while (!czy_wypukly(punkty[otoczka[rozmiar1 - 2]],
                        punkty[otoczka[rozmiar1 - 1]], punkty[i],
                        punkty[otoczka[rozmiar1 - 1]]) &&
           rozmiar1 > 1) {
      --rozmiar1;
    }
    otoczka[rozmiar1] = i;
    ++rozmiar1;
  }
  --i;
  while (!rozw[i]) {
    --i;
  }
  otoczka[rozmiar1] = i;
  --i;
  int rozmiar2 = rozmiar1 + 1;
  for (; i >= 0; --i) {
    if (!rozw[i]) {
      continue;
    }
    while (!czy_wypukly(punkty[otoczka[rozmiar2 - 2]],
                        punkty[otoczka[rozmiar2 - 1]], punkty[i],
                        punkty[otoczka[rozmiar2 - 1]]) &&
           rozmiar2 > rozmiar1) {
      --rozmiar2;
    }
    otoczka[rozmiar2] = i;
    ++rozmiar2;
  }
  return rozmiar2;
}
double dist(pair<int, int> a, pair<int, int> b) {
  return sqrt((a.first - b.first) * (a.first - b.first) +
              (a.second - b.second) * (a.second - b.second));
}
double maxi(double a, double b) { return (a > b ? a : b); }
double najdalsza_para_punktow() {
  int rozmiar = znajdz_otoczke();
  double naj = 0;
  if (rozmiar < 2) {
    return naj;
  }
  int indeks = 1;
  for (int i = 0; i < rozmiar - 1; ++i) {
    while (czy_wypukly(punkty[otoczka[i + 1]], punkty[otoczka[i]],
                       punkty[otoczka[indeks + 1]], punkty[otoczka[indeks]])) {
      ++indeks;
      indeks %= rozmiar - 1;
    }
    naj = maxi(naj, dist(punkty[otoczka[i]], punkty[otoczka[indeks]]));
    naj = maxi(naj, dist(punkty[otoczka[i + 1]], punkty[otoczka[indeks]]));
    naj = maxi(naj, dist(punkty[otoczka[i]], punkty[otoczka[indeks + 1]]));
    naj = maxi(naj, dist(punkty[otoczka[i + 1]], punkty[otoczka[indeks + 1]]));
  }
  return naj;
}
int main() {
  wczytaj_punkty();
  sort(punkty, punkty + N);
  double p = 0, q = 30000, naj;
  while (q - p > 0.0000005) {
    double s = (p + q) / 2;
    for (int i = 0; i < N; ++i) {
      for (int j = 0; j < N; ++j) {
        if (dist(punkty[i], punkty[j]) > s) {
          rozw[j] = 1;
        }
      }
      naj = najdalsza_para_punktow();
      for (int i = 0; i < N; ++i) {
        rozw[i] = 0;
      }
      if (naj >= s) {
        flaga = 1;
        break;
      }
    }
    if (flaga) {
      p = s;
    } else {
      q = s - 0.00000025;
    }
    flaga = 0;
  }
  printf("%.20lf", p / 2);
  return 0;
}