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>
#include     <algorithm>
#include     <vector>
#include     <set>
#include     <map>
#define      boost          ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define      test           int t; cin>>t; while(t-->0)
#define      pb             push_back
#define      pp             pop_back
#define      int            long long
#define      mk             make_pair
#define      fo(i,n)        for(ll i=0;i<n;i++)
#define      fo1(i,n)       for(ll i=1;i<=n;i++)
#define      ll             long long
#define      debug(i,x)     cout<<"Case #"<<i<<": "<<x<<"\n"
#define      debug1(i,x,y)  cout<<"Case #"<<i<<": "<<x<<" "<<y<<"\n"
#define      mx             1000000007
#define      pi             pair<int,int>
#define      pl             pair<long long,long long>
#define      fi             first
#define      se             second
// "     !       " //
using namespace std;

ll power(ll a,ll b){
if(b==0) return 1;
if(b%2==1) return (a*power(a*a,b/2))%mx; 
return power((a*a)%mx,b/2);
}
int gcd(int a, int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b); 
     
}
void solve(){
 int n,k;
 cin>>n>>k;
 int p[k+1],an[k+1],a[n+1];
//  fo1(i,k) a[i]=i;
 int b=k;
//  for(int i=k+1;i<=n;i++){
     
//      a[i]=--k;
//  }
 fo1(i,2*k-n-1) p[i]=i;
 for(int i=2*k-n;i<=k;i++){ 
     p[i]=b;
     b--;
 }
//  fo1(i,n){
     
//      an[i]=p[a[i]];
//  }
// for(int x:p) cout<<x<<" ";
//   cout<<"\n";
  fo1(i,k) {
      
      cout<<p[i]<<" ";
  
  }
   cout<<"\n";
}
signed main(){
    boost
     test{
     solve();
}
}