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:
// In The Name Of God

/*  ki oomade?!  */
#include <bits/stdc++.h>

typedef long long ll;
#define pb push_back
#define S second
#define F first

using namespace std;
const int N = 1e6 + 10;
const ll inf = 1e18;

ll q, n, a[N], ans;
string s, t;
int main()
{
	ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	
  cin >> q;
  while(q --){
    ll k;
    cin >> n >> k;
    vector <ll> v;
    ll l = 1, r = k - 1;
    while(v.size() < k){
      if(v.size() == k - (n - k) - 1) v.pb(k);
      else if(v.size() < k - (n - k) - 1) v.pb(l), l ++;
      else if(v.size() >= k - (n - k)) v.pb(r), r --;
    }
    for(auto x : v) cout << x << ' ';
      cout << endl;
  }	
}
/*
 _____       ____                  _       _ _     _          ____           
|_   _|_ _  / ___| _ __   ___  ___(_) __ _| (_)___| |_ ___   |  _ \ __ _ ___ 
  | |/ _` | \___ \| '_ \ / _ \/ __| |/ _` | | / __| __/ _ \  | |_) / _` / __|
  | | (_| |  ___) | |_) |  __/ (__| | (_| | | \__ \ || (_) | |  __/ (_| \__ \
  |_|\__,_| |____/| .__/ \___|\___|_|\__,_|_|_|___/\__\___/  |_|   \__,_|___/
                  |_|                                                        
 _   _             _                          _                              
| \ | | __ _  __ _(_)_ __ __ _ _ __ ___      / \   _ __ ___   ___  _ __ ___  
|  \| |/ _` |/ _` | | '__/ _` | '_ ` _ \    / _ \ | '__/ _ \ / _ \| '_ ` _ \ 
| |\  | (_| | (_| | | | | (_| | | | | | |  / ___ \| | | (_) | (_) | | | | | |
|_| \_|\__,_|\__, |_|_|  \__,_|_| |_| |_| /_/   \_\_|  \___/ \___/|_| |_| |_|
             |___/                                                           
 _   _                _       _       _                     
| \ | | ___ _ __ ___ (_) __ _(_) __ _(_)_ __ __ _ _ __ ___  
|  \| |/ _ \ '_ ` _ \| |/ _` | |/ _` | | '__/ _` | '_ ` _ \ 
| |\  |  __/ | | | | | | (_| | | (_| | | | | (_| | | | | | |
|_| \_|\___|_| |_| |_|_|\__, |_|\__, |_|_|  \__,_|_| |_| |_|
                        |___/   |___/ 
           																	*/