Problem: Sereja placed n points on a plane. Now Sereja wants to place on the plane two straight lines, intersecting at a right angle, so that one of the straight lines intersect the Ox axis at an angle of 45 degrees and the maximum distance from the points to the straight lines were minimum. 

In this problem we consider the distance between points (x1, y1) and (x2, y2) equal |x1 - x2| + |y1 - y2|. The distance between the point and the straight lines is the minimum distance from the point to some point belonging to one of the lines.

Help Sereja, find the maximum distance from the points to the optimally located straight lines.

Input

The first line contains integer n (1 ≤ n ≤ 105). Next n lines contain the coordinates of the lines. The i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109).

Output

In a single line print a real number — the answer to the problem. Your answer will be considered correct iff its absolute or relative error doesn't exceed 10 - 6.

Examples

Input

4
0 0
2 0
0 2
2 2


Output

0.000000000000000


Input

4
1 0
0 1
2 1
1 2


Output

1.000000000000000

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
inline void smin(T &a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, U b) {
  if (a < b) a = b;
}
template <class T>
inline void gn(T &first) {
  char c, sg = 0;
  while (c = getchar(), (c > '9' || c < '0') && c != '-')
    ;
  for ((c == '-' ? sg = 1, c = getchar() : 0), first = 0; c >= '0' && c <= '9';
       c = getchar())
    first = (first << 1) + (first << 3) + c - '0';
  if (sg) first = -first;
}
template <class T>
inline void print(T first) {
  if (first < 0) {
    putchar('-');
    return print(-first);
  }
  if (first < 10) {
    putchar('0' + first);
    return;
  }
  print(first / 10);
  putchar(first % 10 + '0');
}
template <class T, class T1>
inline void gn(T &first, T1 &second) {
  gn(first);
  gn(second);
}
template <class T, class T1, class T2>
inline void gn(T &first, T1 &second, T2 &z) {
  gn(first);
  gn(second);
  gn(z);
}
template <class T>
inline void println(T first) {
  print(first);
  putchar('\n');
}
template <class T>
inline void printsp(T first) {
  print(first);
  putchar(' ');
}
template <class T1, class T2>
inline void print(T1 x1, T2 x2) {
  printsp(x1), println(x2);
}
template <class T1, class T2, class T3>
inline void print(T1 x1, T2 x2, T3 x3) {
  printsp(x1), printsp(x2), println(x3);
}
template <class T1, class T2, class T3, class T4>
inline void print(T1 x1, T2 x2, T3 x3, T4 x4) {
  printsp(x1), printsp(x2), printsp(x3), println(x4);
}
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
struct point {
  long long first, second;
  point(long long first = 0, long long second = 0)
      : first(first), second(second) {}
  point operator-(point a) { return point(first - a.first, second - a.second); }
  point operator+(point a) { return point(first + a.first, second + a.second); }
  point operator*(int a) { return point(first * a, second * a); }
  point conj() { return point(-second, first); }
  long long operator^(point a) {
    return (long long)first * a.second - (long long)second * a.first;
  }
  long long operator*(point a) {
    return (long long)first * a.first + (long long)second * a.second;
  }
  double nor() { return sqrt(*this * *this); }
  bool operator<(const point &a) const {
    if (first != a.first) return first < a.first;
    return second < a.second;
  }
  bool operator>(const point &a) const {
    if (first != a.first) return first > a.first;
    return second > a.second;
  }
  bool operator==(const point &a) const {
    if (first != a.first) return 0;
    return second == a.second;
  }
  void input() { gn(first, second); }
} p[100111];
long long lmx[100111], lmn[100111], rmx[100111], rmn[100111];
int n;
int ok(double mid) {
  if (p[n].first - p[1].first < 2 * mid + 1e-10) return 1;
  int pos = 1;
  for (int i = 1; i <= n; i++) {
    while (pos <= n and p[pos].first - p[i].first <= 2 * mid) pos++;
    long long MX = max(lmx[i - 1], rmx[pos]);
    long long MN = min(lmn[i - 1], rmn[pos]);
    if (MX - MN < 2 * mid + 1e-10) return 1;
  }
  return 0;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    point a;
    a.input();
    p[i] = point(a.first + a.second, a.first - a.second);
  }
  sort(p + 1, p + n + 1);
  lmx[0] = -0x3f3f3f3f3f3f3f3fLL;
  lmn[0] = 0x3f3f3f3f3f3f3f3fLL;
  for (int i = 1; i <= n; i++) {
    lmx[i] = max(lmx[i - 1], p[i].second);
    lmn[i] = min(lmn[i - 1], p[i].second);
  }
  rmx[n + 1] = -0x3f3f3f3f3f3f3f3fLL;
  rmn[n + 1] = 0x3f3f3f3f3f3f3f3fLL;
  for (int i = n; i >= 1; i--) {
    rmx[i] = max(rmx[i + 1], p[i].second);
    rmn[i] = min(rmn[i + 1], p[i].second);
  }
  sort(p + 1, p + n + 1);
  int step = 100;
  double st = 0, ed = 1e10;
  while (step--) {
    double mid = (st + ed) / 2;
    if (ok(mid))
      ed = mid;
    else
      st = mid;
  }
  printf("%.10lf\n", ed);
  return 0;
}