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:
//package learning;

import java.util.*;
import java.io.*;
import java.lang.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class NitsLocal {
    static ArrayList<String> s1;
    static boolean[] prime;
    static int n = (int)1e7;
    
     static void sieve() {
Arrays.fill(prime	, true);
prime[0] = prime[1] = false;
for(int i = 2 ; i * i <= n ; ++i) {
if(prime[i]) {
for(int k = i * i; k<= n ; k+=i) {
prime[k] = false;
}
}
}

}
    public static void main(String[] args) {
        InputReader sc = new InputReader(System.in);
        
           prime  = new boolean[n + 1];
sieve();
                prime[1] = false;
               int n = sc.ni();
               int [][]ans = new int[n][n];
               int c = 1;
               for(int i=0;i<n;i++)
               {
                   for(int j=0;j<n;j++)
                   {
                       ans[j][i] = c;
                       if(i%2==1)
                       c--;
                       else
                           c++;
                   }
                      if(i%2==0)
                       c += (n-1);
                      else
                          c += (n+1);
               }
              for(int i=0;i<n;i++)
              {
                  for(int j=0;j<n;j++)
                  {
                      w.print(ans[i][j] + " ");
                  }
                  w.println();
              }
               
              
             
                
      
       w.close();
       
        
    }
    public static long pow(long i, long j) {
		long res=1L;
		for(int a=0;a<j;a++) res*=i;
		return res;
	}
    public static long powmod(long i, long j,long MOD) {
		if(j==0) return 1L;
		if(j%2==0) {
			long temp=powmod(i,j/2,MOD);
			return (temp*temp)%MOD;
		}
		return (i*powmod(i,j-1,MOD))%MOD;
	}
    static HashSet<Integer> fac;
    public static void primeFactors(int n) 
    { 
        // Print the number of 2s that divide n 
        int t = 0;
        while (n%2==0) 
        { 
            fac.add(2);
            n /= 2; 
        } 
  
        // n must be odd at this point.  So we can 
        // skip one element (Note i = i +2) 
        for (int i = 3; i*i <= n; i+= 2) 
        { 
            // While i divides n, print i and divide n 
            while (n%i == 0) 
            { 
                fac.add(i);
                n /= i; 
            } 
        } 
  
        // This condition is to handle the case whien 
        // n is a prime number greater than 2 
        if (n > 2) 
            fac.add(n);
    } 
     static long modexp(long x,long n,long M)
{
    long power = n;
    long result=1;
    x = x%M;
    while(power>0)
    {
        if(power % 2 ==1)
            result=(result * x)%M;
        x=(x*x)%M;
        power = power/2;
    }
    return result;
}
    static int gcd(int a,int b)
    {
        if(a==0)
            return b;
        else
            return gcd(b%a,a);
    }
    static class Temp{
        int a;
        int b;
        int d;
        Temp(int a,int b,int d)
        {
            this.a = a;
            this.b = b;
            this.d = d;
        }
    }
    static class Query{
        int p;
        int index;
        int value;
        public Query(int p,int index,int value)
        {
            this.p = p;
            this.index = index;
            this.value = value;
            
        }
    }
    static int nextPowerOf2(int n) 
    { 
        int count = 0; 
  
        // First n in the below  
        // condition is for the  
        // case where n is 0 
        if (n > 0 && (n & (n - 1)) == 0) 
            return n; 
  
        while(n != 0) 
        { 
            n >>= 1; 
            count += 1; 
        } 
  
        return 1 << count; 
    } 
    
   static long sum1(int t1,int t2,int x,int []t)
   {
       int mid = (t2-t1+1)/2;
  if(t1==t2)
      return 0;
    
   else
   return sum1(t1,mid-1,x,t) + sum1(mid,t2,x,t);
    
     
   }
  
   
    static String replace(String s,int a,int n)
    {
        char []c = s.toCharArray();
        for(int i=1;i<n;i+=2)
        {
            int num = (int) (c[i] - 48);
            num += a;
            num%=10;
            c[i] = (char) (num+48);
        }
        return new String(c);
    }
    static String move(String s,int h,int n)
    {
        h%=n;
        char []c = s.toCharArray();
        char []temp = new char[n];
        for(int i=0;i<n;i++)
        {
            temp[(i+h)%n] = c[i];
        }
        return new String(temp);
    }
    
     public static int ip(String s){
return Integer.parseInt(s);
}
     static class multipliers implements Comparator<Long>{
  
         
  public int compare(Long a,Long b) {
   if(a<b)
    return 1;
   else if(b<a)
    return -1;
   else
    return 0;
  }
 }
     static class multipliers1 implements Comparator<Student>{
  
  public int compare(Student a,Student b) {
   if(a.y<b.y)
    return 1;
   else if(b.y<a.y)
    return -1;
   else
   {
       if(a.id < b.id)
           return 1;
       else if(b.id<a.id)
           return -1;
       else
           return 0;
           
    //return 0;
   }
  }
 }
    // Java program to generate power set in 
// lexicographic order. 

    
    
    
      static class InputReader {
 
private final InputStream stream;
private final byte[] buf = new byte[8192];
private int curChar, snumChars;
 
public InputReader(InputStream st) {
this.stream = st;
}
 
public int read() {
if (snumChars == -1)
throw new InputMismatchException();
if (curChar >= snumChars) {
curChar = 0;
try {
snumChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (snumChars <= 0)
return -1;
}
return buf[curChar++];
}
 
public int ni() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
 
public long nl() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
 
public int[] nia(int n) {
int a[] = new int[n];
for (int i = 0; i < n; i++) {
a[i] = ni();
}
return a;
}
 
public String rs() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
 
public String nextLine() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isEndOfLine(c));
return res.toString();
}
 
public boolean isSpaceChar(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
 
private boolean isEndOfLine(int c) {
return c == '\n' || c == '\r' || c == -1;
}
 
}





   
static PrintWriter w = new PrintWriter(System.out);
                  static class Student
    {
        int id;
        //int x;
       
        int y;
        //long z;
        
        Student(int id,int y)
        {
            this.id = id;
            //this.x = x;
            //this.s = s;
            this.y = y;
           // this.z = z;
            
            
            
        }
        
    }
}