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.util.*;
import java.io.*;
import java.math.BigInteger;

public class Main
{
	static final long mod=(int)1e9+7;
	public static void main(String[] args) throws Exception
	{
		FastReader in=new FastReader();
		PrintWriter pw=new PrintWriter(System.out);
		int n=in.nextInt();
		int[][] mat=new int[n][n];
		int a=0;
		for(int i=0;i<n;i+=4)
		{
			for(int j=0;j<n;j+=4)
			{
				for(int i1=i;i1<i+4;i1++)
					for(int j1=j;j1<j+4;j1++)
						mat[i1][j1]=a++;
			}
		}
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<n;j++)
				pw.print(mat[i][j]+" ");
			pw.println();
		}
		pw.flush();
	}
}

class FastReader
{
	BufferedReader br;
	StringTokenizer st;

	public FastReader()
	{
	    br=new BufferedReader(new InputStreamReader(System.in));
	}

	public String next() throws IOException
	{
	    if(st==null || !st.hasMoreElements())
	    {
	        st=new StringTokenizer(br.readLine());
	    }
	    return st.nextToken();
	}

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

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