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 CPP):

Solution:
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdio>

using namespace std;

const int N = 1e6+10;
int a[N];
int b[N];
int p[N],vis[N];
int q;
int n,k;
int main()
{
	cin>>q;
	while(q--)
	{
		memset(vis,0,sizeof(vis));
		cin>>n>>k;
		for(int i = 0, j = k-1; i<n; i++)
		{
			if(i+1<=k)
				a[i] = i+1;
			else
				a[i] = j--;
		}
		int t = k;
		for(int i=0,j=n-1;i<n;i++){
			if(vis[i]) continue;
			if(a[i]!=a[j])  b[i]=a[i];
			else {
				b[i]=b[j]=t;
				vis[j]=1;
				t--;
				j--;
			}
		}
		for(int i = 0; i<n ;i++)
			p[a[i]] = b[i];
		for(int i = 1; i<=k; i++)
		    cout<<p[i]<<" ";
		    cout<<endl;
	}
	return 0;
}