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:
"""
ID: happyn61
LANG: PYTHON3
PROB: loan
"""

#from collections import defaultdict
import sys
import heapq
from collections import deque

#fin = open ('loan.in', 'r')
#fout = open ('loan.out', 'w')
#print(dic["4734"])
def find(parent,i):


    if parent[i] != i: 
        parent[i]=find(parent,parent[i]) 
    return parent[i] 

        # A utility function to do union of two subsets 
def union(parent,rank,xx,yy): 
    x=find(parent,xx)
    y=find(parent,yy)
    if rank[x]>rank[y]:
        parent[y]=x
    elif rank[y]>rank[x]:
        parent[x]=y
    else:
        parent[y]=x
        rank[x]+=1
ans=0

#NQ=sys.stdin.readline().strip().split()

n=int(sys.stdin.readline().strip())
#N1=int(NQ[0])
#N2=int(NQ[1])
#N3=int(NQ[1])

for j in range(n):
    AB=sys.stdin.readline().strip().split()
    n=int(AB[0])
    m=int(AB[1])
    l=[str(i+1) for i in range(m)]
    if 2*m-1-n==0:
        l.reverse()
        print(" ".join(l))
    else:
        ll=l[2*m-1-n:]
        ll.reverse()
        k=l[:2*m-1-n]+ll
        print(" ".join(k))
    #l=sys.stdin.readline().strip().split()
    #s1=sys.stdin.readline().strip()
    #s2=sys.stdin.readline().strip()
    
        #print(pre,post,ll,rr,m1,m2,pre[ll],post[n-1-rr],post[n-1-rr][2])
    #if F:
    #    print("yes")
    #else:
    #    print("no")
    #return True
    
#for x,y in occupy:
#    l[x][y]="X"
#for ll in l:
#    print("".join(ll))