#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define ull unsigned long long int
#define ld long double
#define endl "\n"
#define intarr(arr, n)  ll arr[n];  for (ll i = 0; i < n; i++)  cin >> arr[i]
#define intvec(vec,n) vector<ll> a(n); for(ll i=0;i<n;i++){ll var;cin>>var;vec.push_back(var);}
#define ff first
#define ss second
#define pb push_back
#define m_p make_pair
#define vec vector<ll>
#define pi pair<ll,ll>
#define rep(i,a,b) for(ll i=a;i<b;i++)
#define rev(i,a,b) for(ll i=a;i>=b;i--)
#define all(v) v.begin(),v.end()
#define mod2 998244353
#define SPEED ios_base :: sync_with_stdio(0), cin.tie(0), cout.tie(0)
#define w ll t;cin>>t;while(t--)

int main() 
{
	SPEED;
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif
	w{
		ll n,k;cin>>n>>k;;
		ll u=2*k-n-1;
		rep(i,1,u+1){
			cout<<i<<" ";
		}
		for(int i=k;i>u;i--)
			cout<<i<<" ";
		cout<<endl;

	}
	return 0;
}


### Prompt

Please formulate a Cpp solution to the following 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.

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define ull unsigned long long int
#define ld long double
#define endl "\n"
#define intarr(arr, n)  ll arr[n];  for (ll i = 0; i < n; i++)  cin >> arr[i]
#define intvec(vec,n) vector<ll> a(n); for(ll i=0;i<n;i++){ll var;cin>>var;vec.push_back(var);}
#define ff first
#define ss second
#define pb push_back
#define m_p make_pair
#define vec vector<ll>
#define pi pair<ll,ll>
#define rep(i,a,b) for(ll i=a;i<b;i++)
#define rev(i,a,b) for(ll i=a;i>=b;i--)
#define all(v) v.begin(),v.end()
#define mod2 998244353
#define SPEED ios_base :: sync_with_stdio(0), cin.tie(0), cout.tie(0)
#define w ll t;cin>>t;while(t--)

int main() 
{
	SPEED;
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif
	w{
		ll n,k;cin>>n>>k;;
		ll u=2*k-n-1;
		rep(i,1,u+1){
			cout<<i<<" ";
		}
		for(int i=k;i>u;i--)
			cout<<i<<" ";
		cout<<endl;

	}
	return 0;
}


```