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 PYTHON):

Solution:
 
# Rhythem Jain
# Date: 14.1.21


from __future__ import division, print_function
import os,sys
from io import BytesIO, IOBase

from collections import deque, Counter, OrderedDict,defaultdict
# from binarytree import tree,heap,bst,Node
# from Queue import LifoQueue
#from heapq import nsmallest, nlargest, heapify,heappop ,heappush, heapreplace
#from math import log,sqrt,factorial,cos,tan,sin,radians
#from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
#from decimal import *
#import threading
from itertools import permutations,combinations
# Copy 2D list  m = [x[:] for x in mark] .. Avoid Using Deepcopy 

if sys.version_info[0] < 3:
    from __builtin__ import xrange as range
    from future_builtins import ascii, filter, hex, map, oct, zip

def ii():  return int(input())
def si():  return input()
def mi():  return map(int,input().strip().split())
def msi(): return map(str,input().strip().split())
def li():  return list(mi())
 
def dmain():
    sys.setrecursionlimit(1000000)
    threading.stack_size(1024000)
    thread = threading.Thread(target=main)
    thread.start()
    
from collections import deque, Counter, OrderedDict,defaultdict
from math import *
from itertools import permutations,combinations
 
abc='abcdefghijklmnopqrstuvwxyz'
abd={'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
mod=1000000007
vow=['a','e','i','o','u','A','E','I','O','U']
dx,dy=[-1,1,0,0],[0,0,1,-1]
def getKey(item): return item[1] 
def sort2(l):return sorted(l, key=getKey,reverse=True)
def d2(n,m,num):return [[num for x in range(m)] for y in range(n)]
def isPowerOfTwo (x): return (x and (not(x & (x - 1))) )
def decimalToBinary(n): return bin(n).replace("0b","")
def ntl(n):return [int(i) for i in str(n)]
def ncr(n,r): return factorial(n)//(factorial(r)*factorial(max(n-r,1)))
 
def ceil(x,y):
    if x%y==0:
        return x//y
    else:
        return x//y+1
 
def powerMod(x,y,p):
    res = 1
    x %= p
    while y > 0:
        if y&1:
            res = (res*x)%p
        y = y>>1
        x = (x*x)%p
    return res

def expo(x,y,m):
    if y==0:
        return 1
    elif y%2==0:
        return expo((x*x)%m,y/2,m)
    else:
        return x*expo((x*x)%m,(y-1)/2,m)%m

def gcd(a, b):
    if a==0:
        return b,0,1
    gc,x1,y1=gcd(b%a,a)
    x=y1-(b//a)*x1
    y=x1 
    return gc,x,y

def sieve(n):
    arr=[True]*(n+1)
    arr[0]=False
    arr[1]=False
    for i in range(4,n+1,2):
        arr[i]=False
    for i in range(3,int(sqrt(n))+1,2):
        if arr[i]==True:
            for j in range(i*i,n+1,i):
                arr[j]=False
    return arr

def modInverse(x,y):
    gc,x1,y1 = gcd(x,y)
    return (y1+y)%y

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 read():
    sys.stdin  = open('input.txt', 'r')  
    sys.stdout = open('output.txt', 'w') 

def main():
    for _ in range(ii()):
        n,k=mi()
        arr=[i for i in range(1,k+1)]
        if n==k:
            print(*arr)
        elif n==2*k-1:
            print(*arr[::-1])
        else:
            val=arr[2*k-n-1:]
            arr=arr[:2*k-n-1]+val[::-1]
            print(*arr)
        


BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

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


def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()


if sys.version_info[0] < 3:
    sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

input = lambda: sys.stdin.readline().rstrip("\r\n")


if __name__ == "__main__":
    main()