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;
long long int powmod(long long int a, int b, int n) {
  long long int rm = 1;
  while (b) {
    if (b % 2) {
      rm = (rm * a) % n;
    }
    a = (a * a) % n;
    b /= 2;
  }
  return rm;
}
int n, x[int(1e5 + 10)], y[int(1e5 + 10)],
    t = 0, a[int(1e5 + 10)], mnv[int(1e5 + 10)] = {0}, mxv[int(1e5 + 10)] = {0},
    dpmnud[int(1e5 + 10)] = {0}, dpmxud[int(1e5 + 10)] = {0},
    dpmndu[int(1e5 + 10)] = {0}, dpmxdu[int(1e5 + 10)] = {0};
map<int, int> xy;
double linedis(int i, int j) {
  i = mnv[i], j = mnv[j];
  double x2 = y[i] - (y[j] - x[j]), x1 = x[i];
  return abs(x2 - x1);
}
double getdis2(int p1, int p2) {
  double x2 = (x[p2] + y[p2]) - y[p1];
  return abs(x[p1] - x2);
}
bool process(int p1, int p2, double dis) {
  int mnp = -1, mxp = -1;
  if (p1 == 1 && p2 == t) return true;
  if (p1 > 1) {
    mnp = dpmnud[p1 - 1], mxp = dpmxud[p1 - 1];
  }
  if (p2 < t) {
    if (mnp == -1) {
      mnp = dpmndu[p2 + 1], mxp = dpmxdu[p2 + 1];
    } else {
      int id2 = dpmndu[p2 + 1];
      if (y[id2] + x[id2] > y[mnp] + x[mnp]) {
        mnp = mnp;
      } else
        mnp = id2;
      id2 = dpmxdu[p2 + 1];
      if (y[id2] + x[id2] < y[mxp] + x[mxp]) {
        mxp = mxp;
      } else
        mxp = id2;
    }
  }
  double dis2 = getdis2(mnp, mxp);
  if (dis >= dis2) return true;
  return false;
}
bool check(double dis) {
  dis *= 2.0;
  int p1 = 1, p2 = 1;
  while (p1 <= t) {
    while (p2 <= t) {
      if (linedis(p1, p2) > dis) {
        break;
      }
      bool ans = process(p1, p2, dis);
      if (ans == true) {
        return true;
      }
      p2++;
      if (p2 > t) break;
    }
    if (p2 > t) break;
    p1++;
    p2 = max(p1, p2);
  }
  return false;
}
int main() {
  cin >> n;
  for (int i = int(1); i <= (int)n; i++) cin >> x[i] >> y[i];
  for (int i = 1; i <= n; ++i) {
    int c = y[i] - x[i];
    if (xy[c] == 0) {
      xy[c] = 1;
      a[++t] = c;
    }
  }
  sort(a + 1, a + t + 1);
  for (int i = 1; i <= t; ++i) xy[a[i]] = t - i + 1;
  for (int i = 1; i <= n; ++i) {
    int id = xy[y[i] - x[i]];
    if (x[i] < x[mnv[id]] || mnv[id] == 0) {
      mnv[id] = i;
    }
    if (x[i] > x[mxv[id]] || mxv[id] == 0) {
      mxv[id] = i;
    }
  }
  for (int i = 1; i <= t; ++i) {
    if (i == 1) {
      dpmnud[i] = mnv[i], dpmxud[i] = mxv[i];
    } else {
      if (y[mnv[i]] + x[mnv[i]] < y[dpmnud[i - 1]] + x[dpmnud[i - 1]]) {
        dpmnud[i] = mnv[i];
      } else
        dpmnud[i] = dpmnud[i - 1];
      if (y[mxv[i]] + x[mxv[i]] > y[dpmxud[i - 1]] + x[dpmxud[i - 1]]) {
        dpmxud[i] = mxv[i];
      } else
        dpmxud[i] = dpmxud[i - 1];
    }
  }
  for (int i = t; i >= 1; i--) {
    if (i == t) {
      dpmndu[i] = mnv[i], dpmxdu[i] = mxv[i];
    } else {
      if (y[mnv[i]] + x[mnv[i]] < y[dpmndu[i + 1]] + x[dpmndu[i + 1]]) {
        dpmndu[i] = mnv[i];
      } else
        dpmndu[i] = dpmndu[i + 1];
      if (y[mxv[i]] + x[mxv[i]] > y[dpmxdu[i + 1]] + x[dpmxdu[i + 1]]) {
        dpmxdu[i] = mxv[i];
      } else
        dpmxdu[i] = dpmxdu[i + 1];
    }
  }
  double it = 0, f = 1e15L, m;
  int iter = 80;
  while (iter--) {
    m = (it + f) / 2.0;
    if (check(m))
      f = m;
    else
      it = m;
  }
  printf("%.10lf\n", f);
  return 0;
}