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:
//@@@@@@@@@@ JAI BAJRANG BALI @@@@@@@@@
 
// COMPILE BY:) NAMAN 
 
#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define maxn 10000001
#define endl "\n"
#define inf INT_MAX
#define vec vector <ll>
#define mps map <ll,ll> 
#define ppl vector<pair<ll,ll>> 
#define fi first
#define si second
#define pr pair<ll,ll> 
#define mvec map<ll,vector <ll>>
#define all(v) v.begin(),v.end()
#define infi LLONG_MAX
#define pi 3.141592653
#define ninf INT_MIN
#define ninfi LLONG_MIN
#define lbn lower_bound
#define ubn upper_bound
#define fr(i,a,b)   for(ll int i=a;i<b;i++)
#define rfr(i,a,b)  for(ll int i=a;i>=b;i--)
#define memset(a,b) memset(a,(b),sizeof(a))

void solve()
{
	ll n,k;
	cin>>n>>k;
	vec ans(k+1);
	ll in;
	if(2*k-n>=1)
	{
		in=2*k-n;
		ans[2*k-n]=k;
	}
	else
	{
		in=1;
		ans[in]=k;
	}
	ll c=1;
	for(ll i=1;i<in;i++)
	{
		ans[i]=c;
		c++;
	}
	 c=k-1;
	for(ll i=in+1;i<=k;i++)
	{
		ans[i]=c;
		c--;
	}
	ll j=k-1;
	for(ll i=1;i<=k;i++)
		cout<<ans[i]<<" ";
	cout<<endl;



		
	



	
	
    
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll t;
    cin>>t;
    while(t--)
    {
         solve();
    }
    return 0;
}