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 M 1000000007
#define U 998244353
#define N 1000005
#define sz(c) (int)c.size()
#define fr first
#define ll long long 
#define ull uint64_t 
#define sc second
#define pb push_back
#define mp make_pair
#define all(a) (a).begin(),(a).end()
#define fi(i,a,n) for(int i=a ; i<n ; i++)
#define fd(i,n,a) for(int i=n-1 ; i>=a ; i--)
#define r0 return 0;
#define ln "\n"
#define INF (int)1e15
#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);   

bool prime[100000005];
void seive()
{
    prime[0]=0;
    prime[1]=0;
    for(int i=2;i<100000005;i++){
        if(prime[i]==0){
            for(int j=i*i;j<100000005;j+=i)
                prime[j]=0;
        }
        else continue;
    }
    
}
//euler totient  
// void etf(){
// 	for(int i=0; i < 10000001; ++i)
// 		phi[i] = i;
// 	for(int i=2; i < 10000001; ++i){
// 		if(phi[i] == i ){
// 			for(int j=1; i*j < 10000001; ++j){
// 				phi[i*j] /= i;
// 				phi[i*j] *= (i-1);
// 			}
// 		}
// 	}
// }

int binexp(int a, int b, int m){
	int result = 1;
	while(b>0){
		if(b & 1){
			result = (result  * 1LL * a) % m;
		}
		a = (a * 1LL * a) % m;
		b >>= 1;
	}
	return result;
}

  
void solve(){
  
	ll n,k;cin>>n>>k;//a b
    vector<ll>v;
    fi(i,1,k+1) v.pb(i);
    ll x=n-k,j=k-1;
    x=k-1-x;
    while(x<j) {swap(v[x],v[j]);
        x++;
        j--;
    }
    fi(i,0,v.size()) cout<<v[i]<<" ";
        cout<<"\n";

}

int32_t main(){
 	fast
	int t;cin>>t;
 	//int t=1;
    while(t--) solve();
    
    int z=0;
    for(int p=1;p<=1000;p++)
    	z++;
	r0;
}