# Magic Grid

Let us define a magic grid to be a square matrix of integers of size $n \times n$, satisfying the following conditions.

• All integers from $0$ to $(n^2 – 1)$ inclusive appear in the matrix exactly once.
• Bitwise XOR of all elements in a row or a column must be the same for each row and column.

You are given an integer $n$ which is a multiple of $4$. Construct a magic grid of size $n \times n$.

Input

The only line of input contains an integer $n$ ($4 \leq n \leq 1000$). It is guaranteed that $n$ is a multiple of $4$.

Output

Print a magic grid, i.e. $n$ lines, the $i$-th of which contains $n$ space-separated integers, representing the $i$-th row of the grid.

If there are multiple answers, print any. We can show that an answer always exists.

Examples

input

4


output

8 9 1 13
3 12 7 5
0 2 4 11
6 10 15 14

input

8


output

19 55 11 39 32 36 4 52
51 7 35 31 12 48 28 20
43 23 59 15 0 8 16 44
3 47 27 63 24 40 60 56
34 38 6 54 17 53 9 37
14 50 30 22 49 5 33 29
2 10 18 46 41 21 57 13
26 42 62 58 1 45 25 61

Note

In the first example, XOR of each row and each column is $13$.

In the second example, XOR of each row and each column is $60$.

Solution:

#include <bits/stdc++.h>

using namespace std;

int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n;
cin >> n;
vector<vector<int>> a(n, vector<int>(n));
int x = 0;
for (int i = 0; i < n; i += 4) {
for (int j = 0; j < n; j += 4) {
for (int ii = 0; ii < 4; ii++) {
for (int jj = 0; jj < 4; jj++) {
a[i + ii][j + jj] = x++;
}
}
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (j > 0) cout << " ";
cout << a[i][j];
}
cout << '\n';
}
return 0;
}