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;
ll z=1000000007;
ll fact[400005];
void buildfact()
{
    fact[0]=1;
    for(ll i=1;i<400005;i++)
    {
        fact[i]=(fact[i-1]*i)%z;
    }
}
ll power(ll a,ll p)
{
    if(p==0)
        return 1;
    if(p%2==0)
        return power((a*a)%z,p/2);
    return (a*power(a,p-1))%z;
}
ll ncr(ll n,ll r)
{
    if(n<r)
        return 0;
    ll num=fact[n];
    ll deno=(fact[r]*fact[n-r])%z;
    deno=power(deno,z-2);
    ll ans=(num*deno)%z;
    return ans;
}
//////////////////////////////////////****************//////////////////////////////////////
ll maxp(ll a,ll b)
{
    if(a>b)
        return a;
    return b;
}
void solve(int z)
{
 ll n,k;
 cin>>n>>k;
 if(n==k)
 {
     for(int i=1;i<=n;i++)
        cout<<i<<" ";
    cout<<endl;
     return;
 }
 int temp=k-(n-k+1);
 for(int i=1;i<=temp;i++)
    cout<<i<<" ";
 int t=k;
 while(t>temp)
 {
     cout<<t<<" ";
     t--;
 }
 cout<<endl;
 return;
}
int main()
{
    ll t;
    t=1;
    cin>>t;
   //buildfact();
   int i=1;
    while(i<=t)
    {
        solve(i);
        i++;
    }
    return 0;
}