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<iostream>
#include<stdio.h>
#include<deque>
#include<vector>
#include<map>
#include<math.h>
#include<algorithm>
#include<queue>
#include<bits/stdc++.h>

using namespace std;
#define lli long long int
#define ull unsigned long long int
#define ff first
#define MOD 1000000007
#define ss second
#define print(containeer) \
for(auto iterator:containeer)cout<<iterator<<" ";cout<<endl;
#define PI 3.14159265
#define N 300
struct node{

lli ind, x, y;
};
bool comp( struct node n1 , struct node n2)
{return n1.x<n2.x;}
lli bs(vector<struct node >& v, lli k)
{


    if(v[0].x>=k)
        return -1;
    lli left=0, right=v.size()-1;
    lli ans;
    while(left<=right)
    {

        lli mid=left+(right-left)/2;
        if(v[mid].x<k)
        {
            ans=mid;
            left=mid+1;
        }
        else
            right=mid-1;
    }
    return ans;
}
void solve()
{

    lli a,b,c,d,i,j,k,l;
   cin>>a>>b;
   lli a1[b];
   k=a-b+1;
   for(i=1;i<=b;i++)
    a1[i-1]=i;
deque<lli>v;
i=b-1;
while(k)
{
    v.push_back(a1[i]);
    i--;
    k--;

}
//cout<<v.size()<<endl;
i=0;
while(1)
{
    if(i+v.size()>=b)
        break;
    cout<<a1[i]<<" ";
i++;

}
if(v.size())
   print(v);



}
int main()
{

    lli a=1;
    cin>>a;
    while(a--)solve();
}