Problem: Gerald has n younger brothers and their number happens to be even. One day he bought n2 candy bags. One bag has one candy, one bag has two candies, one bag has three candies and so on. In fact, for each integer k from 1 to n2 he has exactly one bag with k candies. 

Help him give n bags of candies to each brother so that all brothers got the same number of candies.

Input

The single line contains a single integer n (n is even, 2 ≤ n ≤ 100) — the number of Gerald's brothers.

Output

Let's assume that Gerald indexes his brothers with numbers from 1 to n. You need to print n lines, on the i-th line print n integers — the numbers of candies in the bags for the i-th brother. Naturally, all these numbers should be distinct and be within limits from 1 to n2. You can print the numbers in the lines in any order. 

It is guaranteed that the solution exists at the given limits.

Examples

Input

2


Output

1 4
2 3

Note

The sample shows Gerald's actions if he has two brothers. In this case, his bags contain 1, 2, 3 and 4 candies. He can give the bags with 1 and 4 candies to one brother and the bags with 2 and 3 to the other brother.

Solution (in JAVA):

Solution:
/* package codechef; // don't place package name! */

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

/* Name of the class has to be "Main" only if the class is public. */
public class Codechef
{
    public static class FastIO
	{
		BufferedReader br;
		BufferedWriter bw, be;
		StringTokenizer st;

		public FastIO()
		{
			br = new BufferedReader(new InputStreamReader(System.in));
			bw = new BufferedWriter(new OutputStreamWriter(System.out));
			be = new BufferedWriter(new OutputStreamWriter(System.err));
			st = new StringTokenizer("");
		}

		private void read() throws IOException
		{
			st = new StringTokenizer(br.readLine());
		}

		public String ns() throws IOException
		{
			while(!st.hasMoreTokens())
				read();
			return st.nextToken();
		}

		public int ni() throws IOException
		{
			return Integer.parseInt(ns());
		}

		public long nl() throws IOException
		{
			return Long.parseLong(ns());
		}

		public float nf() throws IOException
		{
			return Float.parseFloat(ns());
		}

		public double nd() throws IOException
		{
			return Double.parseDouble(ns());
		}

		public char nc() throws IOException
		{
			return ns().charAt(0);
		}

		public int[] nia(int n) throws IOException
		{
			int[] a = new int[n];
			for(int i = 0; i < n; i++)
				a[i] = ni();

			return a;
		}

		public long[] nla(int n) throws IOException
		{
			long[] a = new long[n];
			for(int i = 0; i < n; i++)
				a[i] = nl();

			return a;
		}

		public void out(String s) throws IOException
		{
			bw.write(s);
		}

		public void flush() throws IOException
		{
			bw.flush();
			be.flush();
		}

		public void err(String s) throws IOException
		{
			be.write(s);
		}
	}
	 static boolean isPrime(int n)
    {
     // Check if number is less than
       // equal to 1
        if (n <= 1)
            return false;
        // Check if number is 2
        else if (n == 2)
            return true;
        // Check if n is a multiple of 2
        else if (n % 2 == 0)
            return false;
        // If not, then just check the odds
        for (int i = 3; i <= Math.sqrt(n); i += 2) 
        {
            if (n % i == 0)
                return false;
        }
        return true;
    }
     static int gcd(int a, int b)
    {
        if (a == 0)
            return b; 
        return gcd(b % a, a); 
    }
     
    // method to return LCM of two numbers
    static int lcm(int a, int b)
    {
        return (a / gcd(a, b)) * b;
    }
    
    static int smallestDivisor(int n)  
{  
    // if divisible by 2  
    if (n % 2 == 0)  
        return 2;  
  
    // iterate from 3 to sqrt(n)  
    for (int i = 3; i * i <= n; i += 2) {  
        if (n % i == 0)  
            return i;  
    }  
  
    return n;  
}  

	static FastIO f;
	
	public static void main(String args[]) throws IOException
	{
		f = new FastIO();

		int n;
		 int x=1;int i,j;
		n = f.ni();int a[][]=new int[n+1][n+1];
		for( i=1;i<=n;i++)
		{  for( j=1;j<=n/2;j++){
		    a[i][j]=x;
		    a[i][n-(j-1)]=(n*n)-(x-1);
		    x++;
		}
		}
		
	
			for( i=1;i<=n;i++)	{
			    for(j=1;j<=n;j++){
			        f.out(a[i][j]+" ");
			    }f.out("\n");
			    }
			
			
		f.flush();
	}
}