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:
from math import *
from sys import *
from heapq import *
from collections import defaultdict
from itertools import permutations
import os, sys
from io import IOBase, BytesIO
M=10**9+7

def expi(bas, exp): 
    t = 1; 
    while(exp > 0):  
  
        # for cases where exponent 
        # is not an even value 
        if (exp % 2 != 0): 
            t = (t * bas) % N; 
  
        bas = (bas * bas) % N; 
        exp = int(exp / 2); 
    return t % N; 


def graph_as_tree_1_root():
    n=int(input())
    dict=defaultdict(list)
    for _ in range(n-1):
        x,y=list(map(int,input().split()))
        dict[x].append(y)
        dict[y].append(x)
    tree=[[1]]
    temp=[1]
    vall=0
    seen=set([i for i in range(2,n+1)])
    lol=[]
    while len(seen)>0:
        #print(seen)
        lol=list(seen)
        tt=[]
        for x in temp:
            aa=[]
            for val in dict[x]:
                if val in seen:
                    aa+=[val]
                    seen.remove(val)
            dict[x]=aa
            tt+=aa
        tree.append(tt)
        temp=tt
        vall+=1
    for x in lol:
        dict[x]=[]
    
    print(tree,dict)
def pow(a,b):
    res=1
    while b>0:
        if b&1:
            res*=a
        a*=a
        b>>=1
    return res
def powmod(a,b,m):
    res=1
    while b>0:
        if b&1:
            res=((res*a)%m)
        a*=a
        b>>=1
    return res
def inv(a,m):
    return powmod(a,m-2,m)
def alldivisors(n) : 
    list = []  
    arr=[]
    for i in range(1, int(sqrt(n) + 1)) :
        if (n % i == 0) : 
            if (n / i == i) : 
                arr+=[i]
            else :
                arr+=[i]
                list.append(n//i)  
    arr+=list[::-1]
    return arr
def primefactorisation(n):
    potentional_p = 3
    itog_list = defaultdict(int)
    if n % 2 == 0:
        itog_list[2] = 0
    while n % 2 == 0:
        n = n // 2
        itog_list[2] += 1
    while n - 1:
        if potentional_p > (n**0.5):
            itog_list[n] += 1
            return itog_list
        while n % potentional_p == 0:
            n = n // potentional_p
            itog_list[potentional_p] += 1
        potentional_p += 2
    return itog_list



def main():
    n=int(input())
    a=[]
    delta=2*n-1
    cc=1
    for i in range(n):
        temp=[cc]
        prev=cc
        fl=1
        while prev<=n*n:
            if fl==1:
                prev=prev+delta
                fl=-1
            else:
                prev+=(2*n-delta)
                fl=1
            temp.append(prev)
        a.append(temp)
        cc+=1
        delta-=2
    for x in a:
        print(*x[:-1])
        











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
        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()