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 PYTHON3):

Solution:
#Mamma don't raises quitter.................................................
from collections import deque as de
import math
from sys import stdin, stdout
import re
from collections import Counter as cnt
from functools import  reduce

from itertools import groupby as gb
#from fractions import Fraction as fr
from bisect import bisect_left as bl, bisect_right as br

def factors(n):    
    return set(reduce(list.__add__, 
                ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
    
class My_stack():
    def __init__(self):
        self.data = []
    def my_push(self, x):
        return (self.data.append(x))
    def my_pop(self):
        return (self.data.pop())
    
    def my_peak(self):
        return (self.data[-1])
    def my_contains(self, x):
        return (self.data.count(x))
    def my_show_all(self):
        return (self.data)
    def isEmpty(self):
      return len(self.data)==0

arrStack = My_stack() 

#decimal to binary   
def decimalToBinary(n): 
    return bin(n).replace("0b", "")
#binary to decimal
def binarytodecimal(n):
    return int(n,2)

def isPrime(n) : 
	if (n <= 1) : 
		return False
	if (n <= 3) : 
		return True

	
	if (n % 2 == 0 or n % 3 == 0) : 
		return False

	i = 5
	while(i * i <= n) : 
		if (n % i == 0 or n % (i + 2) == 0) : 
			return False
		i = i + 6

	return True

def get_prime_factors(number):
    prime_factors = []
    while number % 2 == 0:
        prime_factors.append(2)
        number = number / 2
    for i in range(3, int(math.sqrt(number)) + 1, 2):
        while number % i == 0:
            prime_factors.append(int(i))
            number = number / i

    if number > 2:
        prime_factors.append(int(number))
    return prime_factors

def get_frequency(list):
    dic={}
    for ele in list:
        if ele in dic:
            dic[ele] += 1
        else:
            dic[ele] = 1
    return dic
def Log2(x): 
    return (math.log10(x) / 
            math.log10(2)); 

 
# Function to get product of digits
def getProduct(n):
 
    product = 1
 
    while (n != 0):
        product = product * (n % 10)
        n = n // 10
 
    return product


#function to find LCM of two numbers
def lcm(x,y):
   lcm = (x*y)//math.gcd(x,y)
   return lcm

def isPowerOfTwo(n): 
    return (math.ceil(Log2(n)) == math.floor(Log2(n))); 
#to check whether the given sorted sequnce is forming an AP or not....
def checkisap(list):
    d=list[1]-list[0]
    for i in range(2,len(list)):
        temp=list[i]-list[i-1]
        if temp !=d:
            return False
    return True
        
    

#ceil  function gives wrong answer after 10^17 so i have to create my own :)
# because i don't want to doubt on my solution of 900-1000 problem set.
def ceildiv(x,y): 
    return (x+y-1)//y 
  
def di():return map(int, input().split())
def ii():return int(input())
def li():return list(map(int, input().split()))
def si():return list(map(str, input()))
def indict():
    dic = {}
    for index, value in enumerate(input().split()):
        dic[int(index)+1] = int(value)
    return dic
def frqdict():
    # by default it is for integer input. :)
    dic={}
    for index, value in enumerate(input().split()):
        if int(value) not in dic:
            dic[int(value)] =1
        else:
            dic[int(value)] +=1
    return dic

#inp = open("input.txt","r")
#out = open("output.txt","w")
#Here we go......................
#practice like your never won
#perform like you never lost
n=ii()
ans=[]
row=[]
j=1
for i in range(1, (n*n)+1):
    if i%n:
        row.append(i)
    else:
        row.append(i)
        if j%2==0:
            row.reverse()
        ans.append(row)
        j+=1
        row=[]
for i in range(n):
    for row in ans:
        print(row[i], end=" ")
    print(end="\n")