#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
using namespace std;
namespace _c {
const double pi = acos(-1.0);
namespace min {
const int i8 = -128;
const int i16 = -32768;
const int i = -2147483647 - 1;
const long long l = -9223372036854775807LL - 1;
}  // namespace min
namespace max {
const int i8 = 127;
const int i16 = 32767;
const int i = 2147483647;
const long long l = 9223372036854775807LL;
}  // namespace max
}  // namespace _c
namespace _f {
template <typename T>
inline const T gcd(T m, T n) {
  while (n != 0) {
    T t = m % n;
    m = n;
    n = t;
  }
  return m;
}
template <typename T>
inline const T max(const T &a, const T &b) {
  return a > b ? a : b;
}
template <typename T>
inline const T min(const T &a, const T &b) {
  return a < b ? a : b;
}
template <typename T>
inline const T abs(const T &a) {
  return a > 0 ? a : -a;
}
template <typename T>
inline T pow(T a, T b) {
  T res = 1;
  while (b > 0) {
    if (b & 1) {
      res = res * a;
    }
    a = a * a;
    b >>= 1;
  }
  return res;
}
template <typename T>
inline T pow(T a, T b, const T &mod) {
  a %= mod;
  T res = 1;
  while (b > 0) {
    if (b & 1) {
      res = res * a % mod;
    }
    a = a * a % mod;
    b >>= 1;
  }
  return res % mod;
}
}  // namespace _f
namespace io {
template <typename T>
inline void read(T &t) {
  register T res = 0, neg = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar()) {
    if (c == '-') {
      neg = -1;
    }
  }
  for (; isdigit(c); c = getchar()) {
    res = res * 10 + c - '0';
  }
  t = res * neg;
}
inline int in() {
  register int T;
  read(T);
  return T;
}
inline long long in_ll() {
  register long long T;
  read(T);
  return T;
}
}  // namespace io
const int MXN = 100010;
long long N, ans;
struct coord {
  long long x, y;
  coord() {}
  coord(long long a, long long b) { x = a, y = b; }
} A[MXN];
bool comp(coord a, coord b) { return a.x < b.x; }
multiset<long long> yy;
long long dify() {
  if (yy.empty()) {
    return 0;
  }
  multiset<long long>::iterator it = yy.end();
  it--;
  return *it - *yy.begin();
}
long long difx(long long i, long long j) {
  if (i >= j) {
    return 0;
  }
  return A[j - 1].x - A[i].x;
}
void solve() {
  ans = 100000000000000000LL;
  for (long long i = 0; i < N; i++) {
    yy.insert(A[i].y);
  }
  long long j = 0;
  for (long long i = 0; i < N; i++) {
    while (j < N && dify() > difx(i, j)) {
      ans = min(ans, max(difx(i, j), dify()));
      yy.erase(yy.find(A[j].y));
      j++;
    }
    ans = min(ans, max(difx(i, j), dify()));
    if (i < j) {
      yy.insert(A[i].y);
    } else {
      j++;
    }
  }
  printf("%.6lf\n", ans / 2.);
}
int main() {
  scanf("%lld", &N);
  for (long long i = 0; i < N; i++) {
    long long x, y;
    scanf("%lld%lld", &x, &y);
    A[i] = coord(x - y, x + y);
  }
  sort(A, A + N, comp);
  solve();
}

### Prompt

Create a solution in cpp for the following 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

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
using namespace std;
namespace _c {
const double pi = acos(-1.0);
namespace min {
const int i8 = -128;
const int i16 = -32768;
const int i = -2147483647 - 1;
const long long l = -9223372036854775807LL - 1;
}  // namespace min
namespace max {
const int i8 = 127;
const int i16 = 32767;
const int i = 2147483647;
const long long l = 9223372036854775807LL;
}  // namespace max
}  // namespace _c
namespace _f {
template <typename T>
inline const T gcd(T m, T n) {
  while (n != 0) {
    T t = m % n;
    m = n;
    n = t;
  }
  return m;
}
template <typename T>
inline const T max(const T &a, const T &b) {
  return a > b ? a : b;
}
template <typename T>
inline const T min(const T &a, const T &b) {
  return a < b ? a : b;
}
template <typename T>
inline const T abs(const T &a) {
  return a > 0 ? a : -a;
}
template <typename T>
inline T pow(T a, T b) {
  T res = 1;
  while (b > 0) {
    if (b & 1) {
      res = res * a;
    }
    a = a * a;
    b >>= 1;
  }
  return res;
}
template <typename T>
inline T pow(T a, T b, const T &mod) {
  a %= mod;
  T res = 1;
  while (b > 0) {
    if (b & 1) {
      res = res * a % mod;
    }
    a = a * a % mod;
    b >>= 1;
  }
  return res % mod;
}
}  // namespace _f
namespace io {
template <typename T>
inline void read(T &t) {
  register T res = 0, neg = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar()) {
    if (c == '-') {
      neg = -1;
    }
  }
  for (; isdigit(c); c = getchar()) {
    res = res * 10 + c - '0';
  }
  t = res * neg;
}
inline int in() {
  register int T;
  read(T);
  return T;
}
inline long long in_ll() {
  register long long T;
  read(T);
  return T;
}
}  // namespace io
const int MXN = 100010;
long long N, ans;
struct coord {
  long long x, y;
  coord() {}
  coord(long long a, long long b) { x = a, y = b; }
} A[MXN];
bool comp(coord a, coord b) { return a.x < b.x; }
multiset<long long> yy;
long long dify() {
  if (yy.empty()) {
    return 0;
  }
  multiset<long long>::iterator it = yy.end();
  it--;
  return *it - *yy.begin();
}
long long difx(long long i, long long j) {
  if (i >= j) {
    return 0;
  }
  return A[j - 1].x - A[i].x;
}
void solve() {
  ans = 100000000000000000LL;
  for (long long i = 0; i < N; i++) {
    yy.insert(A[i].y);
  }
  long long j = 0;
  for (long long i = 0; i < N; i++) {
    while (j < N && dify() > difx(i, j)) {
      ans = min(ans, max(difx(i, j), dify()));
      yy.erase(yy.find(A[j].y));
      j++;
    }
    ans = min(ans, max(difx(i, j), dify()));
    if (i < j) {
      yy.insert(A[i].y);
    } else {
      j++;
    }
  }
  printf("%.6lf\n", ans / 2.);
}
int main() {
  scanf("%lld", &N);
  for (long long i = 0; i < N; i++) {
    long long x, y;
    scanf("%lld%lld", &x, &y);
    A[i] = coord(x - y, x + y);
  }
  sort(A, A + N, comp);
  solve();
}

```