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

  * All integers from 0 to (n^2 - 1) inclusive appear in the matrix exactly once. 
  * [Bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#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 × n.

Input

The only line of input contains an integer n (4 ≤ n ≤ 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.

import java.io.*;
import java.util.*;

public class C {

    public void solve() throws IOException {
        int n = nextInt();
        int[][] a = {
                {8, 9, 1, 13},
                {3, 12, 7, 5},
                {0, 2, 4, 11},
                {6, 10, 15, 14}
        };
        int[][] ans = new int[n][n];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                ans[i][j] = a[i % 4][j % 4] + ((i >= 4 || j >= 4) ? 16 * ((i / 4) * (n/4) + j / 4) : 0);
            }
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                out.print(ans[i][j] + " ");
            }
            out.println();
        }
    }

    public void run() {
        try {
            br = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);

            solve();

            out.close();
        } catch (IOException e) {
            e.printStackTrace();
            System.exit(1);
        }
    }

    BufferedReader br;
    StringTokenizer in;
    PrintWriter out;

    public String nextToken() throws IOException {
        while (in == null || !in.hasMoreTokens()) {
            in = new StringTokenizer(br.readLine());
        }
        return in.nextToken();
    }

    public int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    public double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    public long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    public int[] nextArr(int n) throws IOException {
        int[] res = new int[n];
        for (int i = 0; i < n; i++) {
            res[i] = nextInt();
        }
        return res;
    }

    public static void main(String[] args) throws IOException {
        Locale.setDefault(Locale.US);
        new C().run();
    }
}