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:
def zip_sorted(a,b):

	# sorted  by a
	a,b = zip(*sorted(zip(a,b)))
	# sorted by b
	sorted(zip(a, b), key=lambda x: x[1])

	return a,b

def number_to_list(a):

	b = []

	while a>=1:

		c = a%10

		a = int(a/10)


		b.append(c)

	return list(reversed(b))

def str_list_to_int_list(a):


	a = list(map(int,a))

	return a

def make_1d_array(n,inti= 0):

	a = [inti for _ in range(n)]

	return a

def make_2d_array(n,m,inti= 0):

	a = [[inti for _ in range(m)] for _ in range(n)]

	return a

def make_3d_array(n,m,k,inti= 0):

	a = [[[inti for _ in range(k)] for _ in range(m)] for _ in range(n)]

	return a

def gcd(a,b):

	if(b==0): 
		return a

	else:
		return gcd(b,a%b);



import sys
input = sys.stdin.readline
I = lambda : list(map(int,input().split()))
S = lambda : list(map(str,input()))




t,=I()
for t1 in range(t):
	n,k = I()
	b = [i for i in range(1,k+1)]
	if n==k:
		print(*b)
	else:
		d = b[:k-(n-k+1)]+list(reversed(b[k-(n-k+1):]))	
		print(*d)