Problem: You have a sequence a with n elements 1, 2, 3, ..., k - 1, k, k - 1, k - 2, ..., k - (n - k) (k ≤ n < 2k).

Let's call as inversion in a a pair of indices i < j such that a[i] > a[j].

Suppose, you have some permutation p of size k and you build a sequence b of size n in the following manner: b[i] = p[a[i]].

Your goal is to find such permutation p that the total number of inversions in b doesn't exceed the total number of inversions in a, and b is lexicographically maximum.

Small reminder: the sequence of k integers is called a permutation if it contains all integers from 1 to k exactly once.

Another small reminder: a sequence s is lexicographically smaller than another sequence t, if either s is a prefix of t, or for the first i such that s_i ≠ t_i, s_i < t_i holds (in the first position that these sequences are different, s has smaller number than t).

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.

The first and only line of each test case contains two integers n and k (k ≤ n < 2k; 1 ≤ k ≤ 10^5) — the length of the sequence a and its maximum.

It's guaranteed that the total sum of k over test cases doesn't exceed 10^5.

Output

For each test case, print k integers — the permutation p which maximizes b lexicographically without increasing the total number of inversions.

It can be proven that p exists and is unique.

Example

Input


4
1 1
2 2
3 2
4 3


Output


1 
1 2 
2 1 
1 3 2 

Note

In the first test case, the sequence a = [1], there is only one permutation p = [1].

In the second test case, the sequence a = [1, 2]. There is no inversion in a, so there is only one permutation p = [1, 2] which doesn't increase the number of inversions.

In the third test case, a = [1, 2, 1] and has 1 inversion. If we use p = [2, 1], then b = [p[a[1]], p[a[2]], p[a[3]]] = [2, 1, 2] and also has 1 inversion.

In the fourth test case, a = [1, 2, 3, 2], and since p = [1, 3, 2] then b = [1, 3, 2, 3]. Both a and b have 1 inversion and b is the lexicographically maximum.

Solution (in PYTHON3):

Solution:
import sys
read=sys.stdin.buffer.read;readline=sys.stdin.buffer.readline;input=lambda:sys.stdin.readline().rstrip()
import bisect,string,math,time,functools,random,fractions
from heapq import heappush,heappop,heapify
from collections import deque,defaultdict,Counter
from itertools import permutations,combinations,groupby
rep=range;R=range
def I():return int(input())
def LI():return [int(i) for i in input().split()]
def LI_():return [int(i)-1 for i in input().split()]
def S_():return input()
def IS():return input().split()
def LS():return [i for i in input().split()]
def NI(n):return [int(input()) for i in range(n)]
def NI_(n):return [int(input())-1 for i in range(n)]
def NLI(n):return [[int(i) for i in input().split()] for i in range(n)]
def NLI_(n):return [[int(i)-1 for i in input().split()] for i in range(n)]
def StoLI():return [ord(i)-97 for i in input()]
def ItoS(n):return chr(n+97)
def LtoS(ls):return ''.join([chr(i+97) for i in ls])
def RLI(n=8,a=1,b=10):return [random.randint(a,b)for i in range(n)]
def RI(a=1,b=10):return random.randint(a,b)
def INP():
    N=9
    n=random.randint(1,N)
    A=[random.randint(1,n) for i in range(m)]
    return n,A
def Rtest(T):
    case,err=0,0
    for i in range(T):
        inp=INP()
        a1=naive(*inp)
        a2=solve(*inp)
        if a1!=a2:
            print(inp)
            print('naive',a1)
            print('solve',a2)
            err+=1
        case+=1
    print('Tested',case,'case with',err,'errors')
def GI(V,E,ls=None,Directed=False,index=1):
    org_inp=[];g=[[] for i in range(V)]
    FromStdin=True if ls==None else False
    for i in range(E):
        if FromStdin:
            inp=LI()
            org_inp.append(inp)
        else:
            inp=ls[i]
        if len(inp)==2:a,b=inp;c=1
        else:a,b,c=inp
        if index==1:a-=1;b-=1
        aa=(a,c);bb=(b,c);g[a].append(bb)
        if not Directed:g[b].append(aa)
    return g,org_inp
def GGI(h,w,search=None,replacement_of_found='.',mp_def={'#':1,'.':0},boundary=1):
    #h,w,g,sg=GGI(h,w,search=['S','G'],replacement_of_found='.',mp_def={'#':1,'.':0},boundary=1) # sample usage
    mp=[boundary]*(w+2);found={}
    for i in R(h):
        s=input()
        for char in search:
            if char in s:
                found[char]=((i+1)*(w+2)+s.index(char)+1)
                mp_def[char]=mp_def[replacement_of_found]
        mp+=[boundary]+[mp_def[j] for j in s]+[boundary]
    mp+=[boundary]*(w+2)
    return h+2,w+2,mp,found
def TI(n):return GI(n,n-1)
def accum(ls):
    rt=[0]
    for i in ls:rt+=[rt[-1]+i]
    return rt
def bit_combination(n,base=2):
    rt=[]
    for tb in R(base**n):s=[tb//(base**bt)%base for bt in R(n)];rt+=[s]
    return rt
def gcd(x,y):
    if y==0:return x
    if x%y==0:return y
    while x%y!=0:x,y=y,x%y
    return y
def YN(x):print(['NO','YES'][x])
def Yn(x):print(['No','Yes'][x])
def show(*inp,end='\n'):
    if show_flg:print(*inp,end=end)

mo=10**9+7
#mo=998244353
inf=float('inf')
FourNb=[(-1,0),(1,0),(0,1),(0,-1)];EightNb=[(-1,0),(1,0),(0,1),(0,-1),(1,1),(-1,-1),(1,-1),(-1,1)];compas=dict(zip('WENS',FourNb));cursol=dict(zip('LRUD',FourNb))
alp=[chr(ord('a')+i)for i in range(26)]
#sys.setrecursionlimit(10**9)


class Bit:
    def __init__(self,n):
        self.size=n
        self.m=len(bin(self.size))-2
        self.arr=[0]*(2**self.m+1)
        
    def __str__(self):
        return str(self.arr)
        
    def add(self,i,x):
        k=0
        while i<=self.size:
            k+=1
            self.arr[i]+=x
            i+=i&(-i)
        return
    
    def sum(self,i):
        rt=0
        while i>0:
            rt+=self.arr[i]
            i-=i&(-i)
        return rt
    
    def l_bound(self,w):
        if w<=0:
            return 0
        x=0
        k=2**self.m
        while k>0:
            if x+k<self.size and self.arr[x+k]<w:
                w-=self.arr[x+k]
                x+=k
            k//=2
        return x+1
        
    def u_bound(self,w):
        if w<=0:
            return 0
        x=0
        k=2**self.m
        while k>0:
            if x+k<self.size and self.arr[x+k]<=w:
                w-=self.arr[x+k]
                x+=k
            k//=2
        return x+1
        
class Bit0(Bit):
    def add(self,j,x):
        super().add(j+1,x)
    def l_bound(self,w):
        return max(super().l_bound(w)-1,0)
    def u_bound(self,w):
        return max(super().u_bound(w)-1,0)
 
class Multiset(Bit0):
    def __init__(self,max_v):
        super().__init__(max_v)
    def insert(self,x):
        super().add(x,1)
    def find(self,x):
        return super().l_bound(super().sum(x))
    def __str__(self):
        return str(self.arr)
 
class Comb:
    def __init__(self,n,mo=10**9+7):
        self.fac=[0]*(n+1)
        self.inv=[1]*(n+1)
        self.fac[0]=1
        self.fact(n)
        for i in range(1,n+1):
            self.fac[i]=i*self.fac[i-1]%mo
            self.inv[n]*=i
            self.inv[n]%=mo
        self.inv[n]=pow(self.inv[n],mo-2,mo)
        for i in range(1,n):
            self.inv[n-i]=self.inv[n-i+1]*(n-i+1)%mo
        return
    
    def fact(self,n):
        return self.fac[n]
        
    def invf(self,n):
        return self.inv[n]
 
    def comb(self,x,y):
        if y<0 or y>x:
            return 0
        return self.fac[x]*self.inv[x-y]*self.inv[y]%mo
 

show_flg=False
show_flg=True

def inv(b):
    n=len(b)
    rt=0
    for i in range(n):
        for j in range(i+1,n):
            if b[i]>b[j]:
                rt+=1
    return rt

def naive(n,k):
    a=list(range(1,k+1))+list(range(k-(n-k),k)[::-1])
    a=list(range(k))+list(range(k-(n-k)-1,k-1)[::-1])
    iv=inv(a)
    show('naive','(n,k)=',(n,k),a,iv)
    rt=tuple(range(k))
    for p in permutations(range(k)):
        b=[p[i]for i in a]
        if inv(b)<=iv:
            rt=max(rt,p)
            #show(p,a,b,inv(b))
    show(rt)
    ans=solve(n,k)
    print(*ans)

    
def solve(n,k):
    t=n-k
    return list(range(1,k-t))+list(range(k-t,k+1))[::-1]

for _ in range(I()):
    n,k=LI()
    ans=solve(n,k)
    print(*ans)