Problem: In order to do some research, n^2 labs are built on different heights of a mountain. Let's enumerate them with integers from 1 to n^2, such that the lab with the number 1 is at the lowest place, the lab with the number 2 is at the second-lowest place, …, the lab with the number n^2 is at the highest place.

To transport water between the labs, pipes are built between every pair of labs. A pipe can transport at most one unit of water at a time from the lab with the number u to the lab with the number v if u > v.

Now the labs need to be divided into n groups, each group should contain exactly n labs. The labs from different groups can transport water to each other. The sum of units of water that can be sent from a group A to a group B is equal to the number of pairs of labs (u, v) such that the lab with the number u is from the group A, the lab with the number v is from the group B and u > v. Let's denote this value as f(A,B) (i.e. f(A,B) is the sum of units of water that can be sent from a group A to a group B).

For example, if n=3 and there are 3 groups X, Y and Z: X = \{1, 5, 6\}, Y = \{2, 4, 9\} and Z = \{3, 7, 8\}. In this case, the values of f are equal to:

  * f(X,Y)=4 because of 5 → 2, 5 → 4, 6 → 2, 6 → 4, 
  * f(X,Z)=2 because of 5 → 3, 6 → 3, 
  * f(Y,X)=5 because of 2 → 1, 4 → 1, 9 → 1, 9 → 5, 9 → 6, 
  * f(Y,Z)=4 because of 4 → 3, 9 → 3, 9 → 7, 9 → 8, 
  * f(Z,X)=7 because of 3 → 1, 7 → 1, 7 → 5, 7 → 6, 8 → 1, 8 → 5, 8 → 6, 
  * f(Z,Y)=5 because of 3 → 2, 7 → 2, 7 → 4, 8 → 2, 8 → 4. 



Please, divide labs into n groups with size n, such that the value min f(A,B) over all possible pairs of groups A and B (A ≠ B) is maximal.

In other words, divide labs into n groups with size n, such that minimum number of the sum of units of water that can be transported from a group A to a group B for every pair of different groups A and B (A ≠ B) as big as possible.

Note, that the example above doesn't demonstrate an optimal division, but it demonstrates how to calculate the values f for some division.

If there are many optimal divisions, you can find any.

Input

The only line contains one number n (2 ≤ n ≤ 300).

Output

Output n lines:

In the i-th line print n numbers, the numbers of labs of the i-th group, in any order you want.

If there are multiple answers, that maximize the minimum number of the sum of units of water that can be transported from one group the another, you can print any.

Example

Input


3


Output


2 8 5
9 3 4
7 6 1

Note

In the first test we can divide 9 labs into groups \{2, 8, 5\}, \{9, 3, 4\}, \{7, 6, 1\}.

From the first group to the second group we can transport 4 units of water (8 → 3, 8 → 4, 5 → 3, 5 → 4).

From the first group to the third group we can transport 5 units of water (2 → 1, 8 → 7, 8 → 6, 8 → 1, 5 → 1).

From the second group to the first group we can transport 5 units of water (9 → 2, 9 → 8, 9 → 5, 3 → 2, 4 → 2).

From the second group to the third group we can transport 5 units of water (9 → 7, 9 → 6, 9 → 1, 3 → 1, 4 → 1).

From the third group to the first group we can transport 4 units of water (7 → 2, 7 → 5, 6 → 2, 6 → 5).

From the third group to the second group we can transport 4 units of water (7 → 3, 7 → 4, 6 → 3, 6 → 4).

The minimal number of the sum of units of water, that can be transported from one group to another is equal to 4. It can be proved, that it is impossible to make a better division.

Solution (in JAVA):

Solution:

import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.*;


public class wef {
public static class FastReader {
	BufferedReader br;
	StringTokenizer st;
	//it reads the data about the specified point and divide the data about it ,it is quite fast
	//than using direct 

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

	String next() {
		while (st == null || !st.hasMoreTokens()) {
			try {
				st = new StringTokenizer(br.readLine());
			} catch (Exception r) {
				r.printStackTrace();
			}
		}
		return st.nextToken();
	}

	int nextInt() {
		return Integer.parseInt(next());//converts string to integer
	}

	double nextDouble() {
		return Double.parseDouble(next());
	}

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

	String nextLine() {
		String str = "";
		try {
			str = br.readLine();
		} catch (Exception r) {
			r.printStackTrace();
		}
		return str;
	}
}
static ArrayList<String>list1=new ArrayList<String>();
static void combine(String instr, StringBuffer outstr, int index,int k)
{
	if(outstr.length()==k)
	{
		list1.add(outstr.toString());return;
	}
	if(outstr.toString().length()==0)
	outstr.append(instr.charAt(index));
    for (int i = 0; i < instr.length(); i++)
    {
        outstr.append(instr.charAt(i));
       
        combine(instr, outstr, i + 1,k);
        outstr.deleteCharAt(outstr.length() - 1);
    }
   index++;
} 
static ArrayList<ArrayList<Integer>>l=new ArrayList<>();
static void comb(int n,int k,int ind,ArrayList<Integer>list)
{
	if(k==0)
	{
		l.add(new ArrayList<>(list));

		return;
	}
	
	
	for(int i=ind;i<=n;i++)
	{
		list.add(i);
		comb(n,k-1,ind+1,list);
		
		list.remove(list.size()-1);
		
	}
	
	
	
	
	

}
static long sum(long n)
{
	long sum=0;
	while(n!=0)
	{
		sum+=n%10;
		n/=10;
	}
	return sum;
}


static boolean check(HashMap<Integer,Integer>map)
{
	for(int h:map.values())
		if(h>1)
			return false;
	return true;
}

static class Pair implements Comparable<Pair>{
    int x;int y;
    Pair(int x,int y){
        this.x=x;
        this.y=y;
      //  this.i=i;
    }
	@Override
	public int compareTo(Pair o) {
		// TODO Auto-generated method stub
		return x-o.x;
		
	}
}
static long ncr(long n, int k) 
{ long m=(long)1e9+7;
    long C[] = new long[k + 1]; 
     
    // nC0 is 1 
    C[0] = 1;   
   
    for (int i = 1; i <= n; i++) 
    { 
        // Compute next row of pascal  
        // triangle using the previous row 
        for (int j = Math.min(i, k); j > 0; j--) 
            C[j] = (C[j]%m + C[j-1]%m)%m; 
    } 
    return C[k]%m; 
} 
static boolean isPrime(int n) 
{ 
    // Corner cases 
    if (n <= 1) 
        return false; 
    if (n <= 3) 
        return true; 
  
    // This is checked so  
    // that we can skip 
    // middle five numbers 
    // in below loop 
    if (n % 2 == 0 ||  
        n % 3 == 0) 
        return false; 
  
    for (int i = 5; 
             i * i <= n; i = i + 6) 
        if (n % i == 0 || 
            n % (i + 2) == 0) 
            return false; 
  
    return true; 
} 
static long pow(long a, long n, long mod) {
	//		a %= mod;
	long ret = 1;
	int x = 63 - Long.numberOfLeadingZeros(n);
	for (; x >= 0; x--) {
		ret = ret * ret % mod;
		if (n << 63 - x < 0)
			ret = ret * a % mod;
	}
	return ret;
}



static int maxsum=0;
static long dp[]=new long[100000];
static long recur(int a[],int ind,int tempsum)
 {
	 if(ind>=a.length)
	 {//System.out.println(tempsum);
		 if(maxsum==0)
			 maxsum=tempsum;
		 else
		 maxsum=Math.max(maxsum, tempsum);
		 return maxsum;
	 }
	 if(dp[ind]!=-1)
		 return dp[ind];
	 
	// System.out.println(tempsum);
	 int ct=0;
	 for(int i=ind;i<ind+(a.length-ind)/2;i++)
	 {ct++;
//	 if(ind+ct<a.length)
	 {	 tempsum+=a[i];
		 
		dp[i]= recur(a,ind+(2*ct),tempsum);}
		 
	 }
	 
	 return dp[ind];
 }
 static int gcd(int a,int b)
 {
	 if(b==0)
		 return a;
	 return gcd(b,a%b);
 }
 static int lcm(int a,int b)
 {
	 return (a*b)/gcd(a,b);
 }
 static HashSet<Long> factorize(long n) {
	    HashSet<Long> facts=new HashSet<Long>();
	    for (long i = 2; i * i <= n; ++i) {
	        while (n % i == 0) {
	            facts.add(i);
	            n /= i;
	        }
	    }
	 
	    if (n != 1) {
	        facts.add(n);
	    }
	    return facts;
	}

public static PrintWriter out = new PrintWriter (new BufferedOutputStream(System.out));

	public static void main(String[] args) {
		// TODO Auto-generated method stub
	FastReader in=new FastReader();
	TreeMap<Long,Integer>map=new TreeMap<Long,Integer>();
	ArrayList<Long>list=new ArrayList<Long>();
//	HashSet<Integer>set=new HashSet<Integer>();
	HashSet<Integer>dis=new HashSet<Integer>();
	Deque<Integer>qu=new LinkedList<Integer>();
	
	
	
	int n=in.nextInt();
	
	int a[][]=new int[n][n];
	int no=1;int ct=0;
	int row=0;int col=0;int pt=1;
	while(no<=n*n)
	{
		
		a[row][col]=no;no++;
		row+=pt;
		if(row==n)
			{
			pt*=-1;row+=pt;col++;
			}
		if(row==-1&&col!=0)
		{
			pt*=-1;row+=pt;col++;
		}
	}
	
	for(int i=0;i<n;i++)
	{
		for(int j=0;j<n;j++)
		{
			out.print(a[i][j]+" ");
		}
		out.println();
	}
	out.flush();
	
	
	
	
	
	
	
	//out.flush();
		
	}
}