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.util.*;
import java.io.*;
public class problem2 {
	public static void merge(int arr[], int l, int m, int r) 
    { 
        // Find sizes of two subarrays to be merged 
        int n1 = m - l + 1; 
        int n2 = r - m; 
  
        /* Create temp arrays */
        int L[] = new int [n1]; 
        int R[] = new int [n2]; 
  
        /*Copy data to temp arrays*/
        for (int i=0; i<n1; ++i) 
            L[i] = arr[l + i]; 
        for (int j=0; j<n2; ++j) 
            R[j] = arr[m + 1+ j]; 
  
  
        /* Merge the temp arrays */
  
        // Initial indexes of first and second subarrays 
        int i = 0, j = 0; 
  
        // Initial index of merged subarry array 
        int k = l; 
        while (i < n1 && j < n2) 
        { 
            if (L[i] <= R[j]) 
            { 
                arr[k] = L[i]; 
                i++; 
            } 
            else
            { 
                arr[k] = R[j]; 
                j++; 
            } 
            k++; 
        } 
  
        /* Copy remaining elements of L[] if any */
        while (i < n1) 
        { 
            arr[k] = L[i]; 
            i++; 
            k++; 
        } 
  
        /* Copy remaining elements of R[] if any */
        while (j < n2) 
        { 
            arr[k] = R[j]; 
            j++; 
            k++; 
        } 
    } 
  
    // Main function that sorts arr[l..r] using 
    // merge() 
    public static void sort(int arr[], int l, int r) 
    { 
        if (l < r) 
        { 
            // Find the middle point 
            int m = (l+r)/2; 
  
            // Sort first and second halves 
            sort(arr, l, m); 
            sort(arr , m+1, r); 
  
            // Merge the sorted halves 
            merge(arr, l, m, r); 
        } 
    } 
    public static int M = 1000000007;
   public static int modularExponentiation(int x,int n)
    {
        if(n==0)
            return 1;
        else if(n%2 == 0)        //n is even
            return modularExponentiation((x*x)%M,n/2);
        else                             //n is odd
            return (x*modularExponentiation((x*x)%M,(n-1)/2))%M;

    }
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		int dir = 0;
		int arr[][] = new int[n][n];
		int t = 0;
		int k = 1;
		for(int i = 1;i<=n;i++) {
			if(dir == 0) {
				for(int j = 0;j<n;j++) {
					arr[j][t] = k;
					k++;
				}
				t++;
				dir = 1;
			}
			else if(dir == 1) {
				for(int j = n-1;j>=0;j--) {
					arr[j][t] = k;
					k++;
				}
				t++;
				dir = 0;
			}
		}
		for(int i = 0;i<n;i++) {
			for(int j = 0;j<n;j++) {
				System.out.print(arr[i][j] + " ");
			}
			System.out.println();
		}
    }

}