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:
/**
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⡀⠀⠀⠀⠀⢀⡀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀
⠀⠀⠀⠀⠀⠀⠀⠀⠀⣠⡖⠁⠀⠀⠀⠀⠀⠀⠈⢲⣄⠀⠀⠀⠀⠀⠀⠀⠀⠀
⠀⠀⠀⠀⠀⠀⠀⠀⣼⡏⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢹⣧⠀⠀⠀⠀⠀⠀⠀⠀
⠀⠀⠀⠀⠀⠀⠀⣸⣿⠁⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈⣿⣇⠀⠀⠀⠀⠀⠀⠀
⠀⠀⠀⠀⠀⠀⠀⣿⣿⡇⠀⢀⣀⣤⣤⣤⣤⣀⡀⠀⢸⣿⣿⠀⠀⠀⠀⠀⠀⠀
⠀⠀⠀⠀⠀⠀⠀⢻⣿⣿⣔⢿⡿⠟⠛⠛⠻⢿⡿⣢⣿⣿⡟⠀⠀⠀⠀⠀⠀⠀
⠀⠀⠀⠀⣀⣤⣶⣾⣿⣿⣿⣷⣤⣀⡀⢀⣀⣤⣾⣿⣿⣿⣷⣶⣤⡀⠀⠀⠀⠀
⠀⠀⢠⣾⣿⡿⠿⠿⠿⣿⣿⣿⣿⡿⠏⠻⢿⣿⣿⣿⣿⠿⠿⠿⢿⣿⣷⡀⠀⠀
⠀⢠⡿⠋⠁⠀⠀⢸⣿⡇⠉⠻⣿⠇⠀⠀⠸⣿⡿⠋⢰⣿⡇⠀⠀⠈⠙⢿⡄⠀
⠀⡿⠁⠀⠀⠀⠀⠘⣿⣷⡀⠀⠰⣿⣶⣶⣿⡎⠀⢀⣾⣿⠇⠀⠀⠀⠀⠈⢿⠀
⠀⡇⠀⠀⠀⠀⠀⠀⠹⣿⣷⣄⠀⣿⣿⣿⣿⠀⣠⣾⣿⠏⠀⠀⠀⠀⠀⠀⢸⠀
⠀⠁⠀⠀⠀⠀⠀⠀⠀⠈⠻⢿⢇⣿⣿⣿⣿⡸⣿⠟⠁⠀⠀⠀⠀⠀⠀⠀⠈⠀
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⣼⣿⣿⣿⣿⣧⡀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀
⠀⠀⠀⠐⢤⣀⣀⢀⣀⣠⣴⣿⣿⠿⠋⠙⠿⣿⣿⣦⣄⣀⠀⠀⣀⡠⠂⠀⠀⠀
⠀⠀⠀⠀⠀⠈⠉⠛⠛⠛⠛⠉⠀⠀⠀⠀⠀⠈⠉⠛⠛⠛⠛⠋⠁⠀⠀
 *    author:  Daredevil666
 *    institution: IIT Patna
**/
#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
#define nl cout<<"\n";
#define ll long long int
#define F(i,a,b) for(i=a;i<b;i++)
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
const double PI = 3.14159265358979323846264338327950288419716939937510582097494459230781641;
const ll M=1e16;
const ll MAXN=2020;
ll i,j,mask;
void solve()
{
   ll n,k;
   cin>>n>>k;
   vector<ll> v(n+1);
   F(i,1,k+1)
   {
       v[i]=i;
   }
   ll t=k;
   ll p=n-k;
   F(i,k-p,k+1)
   {
       v[i]=t;
       t--;
   }
   F(i,1,k+1)
   {
       cout<<v[i]<<" ";
   }
}
int main()
{
IOS
/*#ifndef ONLINE_JUDGE
   freopen("input.txt","r",stdin);
   freopen("output.txt","w",stdout);
#*/
ll t,test;
cin>>test;
F(t,1,test+1)
{
    //cout<<"Case #"<<t<<": ";
    solve();
    nl
}
//solve();
return 0;
}