#include<bits/stdc++.h>
#include<iostream>
#include <iomanip>
#include<math.h>
using namespace std;
#define ll long long
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pb push_back
#define fi first
#define se second
#define gap ' '
#define abs(x) ((x)>0?(x):-(x))
#define INF 1e18
#define mod 998244353
#define inf 1e9
#define pi 3.14159265358979
ll powr(ll m,ll n)
{
    if(m==1)
        return 1;
    if(n==0)
        return 1;
    if(n==1)
        return m%mod;
    if(n%2)
    {
        ll te=powr(m,n/2);
        return (((te*te)%mod)*(m%mod))%mod;
    }
    else
    {
        ll te=powr(m,n/2);
        return (te*te)%mod;
    }
}

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    int tc=1;
    cin>>tc;
    int cs=0;

    while(tc--)
    {
        ll n,k;
        cin>>n>>k;
        vector<int> v;
        for(int i=1;i<=k;i++)
            v.pb(i);
            int tm=(2*k)-n-1;
       if(tm>=0) reverse(v.begin()+tm,v.end());
        for(int i=0;i<v.size();i++)
            cout<<v[i]<<gap;
        cout<<endl;


    }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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.

### Response

```cpp
#include<bits/stdc++.h>
#include<iostream>
#include <iomanip>
#include<math.h>
using namespace std;
#define ll long long
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pb push_back
#define fi first
#define se second
#define gap ' '
#define abs(x) ((x)>0?(x):-(x))
#define INF 1e18
#define mod 998244353
#define inf 1e9
#define pi 3.14159265358979
ll powr(ll m,ll n)
{
    if(m==1)
        return 1;
    if(n==0)
        return 1;
    if(n==1)
        return m%mod;
    if(n%2)
    {
        ll te=powr(m,n/2);
        return (((te*te)%mod)*(m%mod))%mod;
    }
    else
    {
        ll te=powr(m,n/2);
        return (te*te)%mod;
    }
}

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    int tc=1;
    cin>>tc;
    int cs=0;

    while(tc--)
    {
        ll n,k;
        cin>>n>>k;
        vector<int> v;
        for(int i=1;i<=k;i++)
            v.pb(i);
            int tm=(2*k)-n-1;
       if(tm>=0) reverse(v.begin()+tm,v.end());
        for(int i=0;i<v.size();i++)
            cout<<v[i]<<gap;
        cout<<endl;


    }
}

```