#include <bits/stdc++.h>
using namespace std;

int main() {
    int64_t n, a, b;
    cin >> n;
    vector<int64_t> x(n), y(n);
    for (auto i = 0; i < n; i++) {
        cin >> a >> b;
        x[i] = a - b;
        y[i] = a + b;
    }
    sort(x.begin(), x.end());
    sort(y.begin(), y.end());
    cout << max(abs(x.back() - x.front()),
                abs(y.back() - y.front()))
         << endl;
    return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

There are N points on the 2D plane, i-th of which is located on (x_i, y_i). There can be multiple points that share the same coordinate. What is the maximum possible Manhattan distance between two distinct points?

Here, the Manhattan distance between two points (x_i, y_i) and (x_j, y_j) is defined by |x_i-x_j| + |y_i-y_j|.

Constraints

* 2 \leq N \leq 2 \times 10^5
* 1 \leq x_i,y_i \leq 10^9
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
x_1 y_1
x_2 y_2
:
x_N y_N


Output

Print the answer.

Examples

Input

3
1 1
2 4
3 2


Output

4


Input

2
1 1
1 1


Output

0

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

int main() {
    int64_t n, a, b;
    cin >> n;
    vector<int64_t> x(n), y(n);
    for (auto i = 0; i < n; i++) {
        cin >> a >> b;
        x[i] = a - b;
        y[i] = a + b;
    }
    sort(x.begin(), x.end());
    sort(y.begin(), y.end());
    cout << max(abs(x.back() - x.front()),
                abs(y.back() - y.front()))
         << endl;
    return 0;
}
```