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
import os
import math
import copy
from bisect import bisect
from io import BytesIO, IOBase
from math import sqrt,floor,factorial,gcd,log,ceil
from collections import deque,Counter,defaultdict
from itertools import permutations,combinations,accumulate

def Int():               return int(sys.stdin.readline())
def Mint():              return map(int,sys.stdin.readline().split())
def Lstr():              return list(sys.stdin.readline().strip())
def Str():               return sys.stdin.readline().strip()
def Mstr():              return map(str,sys.stdin.readline().strip().split())
def List():              return list(map(int,sys.stdin.readline().split()))
def Hash():              return dict()
def Mod():               return 1000000007
def Mat2x2(n):           return [List() for _ in range(n)]
def Lcm(x,y):            return (x*y)//gcd(x,y)
def dtob(n):             return bin(n).replace("0b","")
def btod(n):             return int(n,2) 
def watch(x):            return print(x)
def common(l1, l2):      return set(l1).intersection(l2)
def Most_frequent(list): return max(set(list), key = list.count)
   
def solution():
     for _ in range(Int()):
          n,k=Mint()
          a=[]
          for i in range(1,k+1):
               a.append(i)
          x=k-1     
          for i in range(k+1,n+1):
               a.append(x)
               x-=1
          b=Counter(a)
          ans=[]
          for i in range(n):
               if(b[a[i]]==2):
                    b[a[i]]=1
               else:
                    ans.append(a[i])
          print(*ans)
          
          
if __name__ == "__main__":
     solution()