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 <bits/stdc++.h>
using namespace std ;
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL);
#define time cerr<<"time taken : "<<(float)clock()/CLOCKS_PER_SEC<<" secs"<<endl;
typedef long long ll ;
typedef unsigned long long int ull;
ll MOD = 10000007 ;

void solve() {

  ll n, k ;
  cin >> n >> k  ;

  ll i ;

  vector<ll>arr(n + 1) ;

  for (i = 1 ; i <= k; i++) {
    arr[i] =  i  ;
  }

  ll t = k - 1 ;

  for (i = k + 1; i <= n; i++) {
    arr[i] = t ;
    t-- ;
  }
  // ll j ;

  // ll inv = 0    ;

  // for (i = 0 , j = 1; i < n - 1, j < n; i++, j++) {

  //   if (arr[i] > arr[j] ) {
  //     inv++ ;
  //   }
  // }

  // for (i = 1 ; i <= n; i++) {
  //   cout << arr[i] ;
  // }


  vector<ll>occ(n + 1)  ;
  vector<ll>ans;

  for (i = n; i >= 1; i--) {

    if (occ[arr[i]] == 1) {
      continue ;
    }
    else {

      ans.push_back(arr[i])  ;

      occ[arr[i]] = 1;
    }
  }

  reverse(ans.begin() , ans.end())   ;


  for (i = 0 ; i < ans.size()  ; i++) {
    cout << ans[i] << " " ;
  }

  //cout << inv << "\n" ;

  cout << endl ;



}
int main()
{

  fast ; time ;

#ifndef ONLINE_JUDGE
  freopen("input.txt", "r", stdin);
  freopen("error.txt", "w", stderr);
  freopen("output.txt", "w", stdout);
#endif

  ll t;
  cin >> t;

  while (t--) {
    solve()   ;
  }

  return 0  ;
}