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;

typedef long long ll;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<string> vs;
typedef vector<vs> vvs;
typedef vector<pii> vpii;
typedef set<int> si;
typedef unordered_set<int> usi;
typedef multiset<int> msi;
typedef map<int,int> mi;
typedef unordered_map<int,int> umi;
typedef multimap<int,int> mmi;

#define f1(a,b,c) for(int(a)=(b);(a)<(c);(a)++)
#define f2(a,b,c) for(int(a)=(b);(a)<=(c);(a)++)
#define f3(a,b,c) for(int(a)=(b);(a)>(c);(a)--)
#define f4(a,b,c) for(int(a)=(b);(a)>=(c);(a)--)
#define fi(a,b) for(auto(a)=b.begin();(a)!=b.end();(a)++)
#define pb push_back
#define all(v) v.begin(),v.end()
#define nper next_permutation
#define pper prev_permutation
#define mp make_pair
#define F first
#define S second

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	int t;
	cin>>t;
	f1(i,0,t)
	{
		int n,k,x=1;
		cin>>n>>k;
		f1(j,0,(2*k)-n-1)
		{
			cout<<x<<" ";x++;
		}
		x=k;
		f1(j,0,n-k+1)
		{
			cout<<x<<" ";x--;
		}
		cout<<"\n";
	}
	/*f1(i,0,t)
	{
		cout<<v[i]<<"\n";
	}*/
	return 0;
}