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 PYTHON):

Solution:
mod=10**9+7
#import resource
#resource.setrlimit(resource.RLIMIT_STACK, [0x100000000, resource.RLIM_INFINITY])
#import threading
#threading.stack_size(2**27)
#import sys
#sys.setrecursionlimit(10**6)
#fact=[1]
#for i in range(1,1000001):
#    fact.append((fact[-1]*i)%mod)
#ifact=[0]*1000001
#ifact[1000000]=pow(fact[1000000],mod-2,mod)
#for i in range(1000000,0,-1):
#    ifact[i-1]=(i*ifact[i])%mod
from sys import stdin, stdout
import bisect
from bisect import bisect_left as bl              #c++ lowerbound bl(array,element)
from bisect import bisect_right as br             #c++ upperbound
import itertools
import collections
import math
import heapq
from random import randint as rn
from Queue import Queue as Q
def modinv(n,p):
    return pow(n,p-2,p)
def ncr(n,r,p):                        #for using this uncomment the lines calculating fact and ifact
    t=((fact[n])*((ifact[r]*ifact[n-r])%p))%p
    return t
def ain():                           #takes array as input
    return list(map(int,sin().split()))
def sin():
    return input().strip()
def GCD(x,y):
    while(y):
        x, y = y, x % y
    return x
"""*******************************************************"""
def main():
    n=int(input())
    b=[[] for i in range(n)]
    x=n**2
    t=0
    while(x>0):
        if(t==0):
            for i in range(n):
                b[i].append(x)
                x-=1
            t=1
        else:
            for i in range(n-1,-1,-1):
                b[i].append(x)
                x-=1
            t=0
    for i in range(n):
        for j in range(n):
            print b[i][j],
        print
######## Python 2 and 3 footer by Pajenegod and c1729
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange

import os, sys
from io import IOBase, BytesIO

BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s
    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()

    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s:self.buffer.write(s.encode('ascii'))
            self.read = lambda:self.buffer.read().decode('ascii')
            self.readline = lambda:self.buffer.readline().decode('ascii')

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')

if __name__ == '__main__':
    main()
#threading.Thread(target=main).start()