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;
const int N = 100010;
struct tree {
  int lt, rt;
  long long ma, mi;
} a[N * 4];
struct point {
  long long x, y;
  friend bool operator<(const point &p, const point &q) { return p.x < q.x; }
} c[N];
void init(int lt, int rt, int step) {
  a[step].lt = lt;
  a[step].rt = rt;
  if (lt == rt) {
    a[step].ma = a[step].mi = c[lt].y;
    return;
  }
  int mid = (lt + rt) / 2;
  init(lt, mid, 2 * step);
  init(mid + 1, rt, 2 * step + 1);
  a[step].ma = max(a[2 * step].ma, a[2 * step + 1].ma);
  a[step].mi = min(a[2 * step].mi, a[2 * step + 1].mi);
}
void query(int lt, int rt, int step, long long &ma, long long &mi) {
  if (a[step].lt == lt && a[step].rt == rt) {
    ma = a[step].ma;
    mi = a[step].mi;
    return;
  }
  if (rt <= a[2 * step].rt)
    query(lt, rt, 2 * step, ma, mi);
  else if (lt > a[2 * step].rt)
    query(lt, rt, 2 * step + 1, ma, mi);
  else {
    long long ma1, ma2, mi1, mi2;
    query(lt, a[2 * step].rt, 2 * step, ma1, mi1);
    query(a[2 * step + 1].lt, rt, 2 * step + 1, ma2, mi2);
    ma = max(ma1, ma2);
    mi = min(mi1, mi2);
  }
}
bool check(long long m, int n) {
  for (int i = 1; i <= n; i++) {
    int lt = 1, rt = i, mid, x = -1, y = -1;
    while (lt <= rt) {
      mid = (lt + rt) / 2;
      if (c[mid].x < c[i].x - m) {
        x = mid;
        lt = mid + 1;
      } else
        rt = mid - 1;
    }
    if (i < n) y = i + 1;
    if (x == -1 && y == -1) return true;
    if (x != -1 && y != -1) {
      long long ma1, ma2, ma, mi1, mi2, mi;
      query(1, x, 1, ma1, mi1);
      query(y, n, 1, ma2, mi2);
      ma = max(ma1, ma2);
      mi = min(mi1, mi2);
      if (ma - mi <= m) return true;
    } else if (x == -1) {
      long long ma, mi;
      query(y, n, 1, ma, mi);
      if (ma - mi <= m) return true;
    } else if (y == -1) {
      long long ma, mi;
      query(1, x, 1, ma, mi);
      if (ma - mi <= m) return true;
    }
  }
  return false;
}
int main() {
  int x, y;
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &x, &y);
    c[i].x = x - y;
    c[i].y = x + y;
  }
  sort(c + 1, c + n + 1);
  init(1, n, 1);
  long long lt = 0, rt = 1LL << 32, mid, ans = 0;
  while (lt <= rt) {
    mid = (lt + rt) / 2;
    if (check(mid, n)) {
      rt = mid - 1;
      ans = mid;
    } else
      lt = mid + 1;
  }
  printf("%.20lf\n", ans / 2.0);
  return 0;
}