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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const int maxN = 110000;
const long long inf = 1000000000000000000LL;
int n;
long long x[maxN], y[maxN];
int pos[maxN];
long long tMin[4 * maxN];
long long tMax[4 * maxN];
vector<pair<long long, long long> > p;
void build(int i, int l, int r) {
  if (l == r) {
    tMin[i] = inf;
    tMax[i] = -inf;
    return;
  }
  build(2 * i, l, (l + r) / 2);
  build(2 * i + 1, (l + r) / 2 + 1, r);
  tMin[i] = min(tMin[2 * i], tMin[2 * i + 1]);
  tMax[i] = max(tMax[2 * i], tMax[2 * i + 1]);
}
void update(int i, int l, int r, int k, long long value) {
  if (l == r) {
    tMin[i] = min(tMin[i], value);
    tMax[i] = max(tMax[i], value);
    return;
  }
  if (k > (l + r) / 2) {
    update(2 * i + 1, (l + r) / 2 + 1, r, k, value);
  } else {
    update(2 * i, l, (l + r) / 2, k, value);
  }
  tMin[i] = min(tMin[2 * i], tMin[2 * i + 1]);
  tMax[i] = max(tMax[2 * i], tMax[2 * i + 1]);
}
pair<long long, long long> get(int i, int l, int r, int cl, int cr) {
  if (cl > cr) {
    return make_pair(inf, -inf);
  }
  if (l == cl && r == cr) {
    return make_pair(tMin[i], tMax[i]);
  }
  if (cl > (l + r) / 2) {
    return get(2 * i + 1, (l + r) / 2 + 1, r, cl, cr);
  }
  if (cr <= (l + r) / 2) {
    return get(2 * i, l, (l + r) / 2, cl, cr);
  }
  pair<long long, long long> A = get(2 * i, l, (l + r) / 2, cl, (l + r) / 2);
  pair<long long, long long> B =
      get(2 * i + 1, (l + r) / 2 + 1, r, (l + r) / 2 + 1, cr);
  return make_pair(min(A.first, B.first), max(A.second, B.second));
}
void precalc() {
  vector<long long> a;
  for (int i = 0; i < n; ++i) {
    a.push_back(x[i]);
    p.push_back(make_pair(x[i], y[i]));
  }
  sort(p.begin(), p.end());
  sort(a.begin(), a.end());
  a.resize(unique(a.begin(), a.end()) - a.begin());
  build(1, 0, n - 1);
  for (int i = 0; i < n; ++i) {
    pos[i] = lower_bound(a.begin(), a.end(), p[i].first) - a.begin();
    update(1, 0, n - 1, pos[i], p[i].second);
  }
}
bool check(long long d) {
  for (int i = 0; i < p.size(); ++i) {
    int start =
        upper_bound(p.begin(), p.end(), make_pair(p[i].first + d, inf)) -
        p.begin();
    pair<long long, long long> A = get(1, 0, n - 1, 0, pos[i] - 1);
    pair<long long, long long> B = make_pair(inf, -inf);
    if (start != p.size()) {
      B = get(1, 0, n - 1, pos[start], n - 1);
    }
    long long L = min(A.first, B.first);
    long long R = max(A.second, B.second);
    if (L + d >= R) {
      return true;
    }
  }
  return false;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    x[i] = u + v;
    y[i] = u - v;
  }
  precalc();
  long long left_bound = 0LL, right_bound = 10000000000LL;
  check(0);
  while (right_bound - left_bound > 1LL) {
    long long middle = (left_bound + right_bound) / 2;
    if (check(middle)) {
      right_bound = middle;
    } else {
      left_bound = middle;
    }
  }
  long long res = right_bound;
  if (check(left_bound)) {
    res = left_bound;
  }
  printf("%.10lf\n", (double)(res) / 2.0);
  return 0;
}