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 <boost/multiprecision/cpp_int.hpp> 
// using boost::multiprecision::cpp_int; 
using namespace std;
// #include <ext/pb_ds/assoc_container.hpp> 
// #include <ext/pb_ds/tree_policy.hpp> 
// using namespace __gnu_pbds; 

// #define OST tree < pair <int, int>,  null_type ,  less<pair <int, int>> ,  rb_tree_tag ,  tree_order_statistics_node_update >

#define ll long long 
#define pb push_back
#define ull unsigned long long
#define ff first
#define ss second
#define IOS ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define test int t;cin>>t;while(t--)
#define mp make_pair
#define modInv(n) po(n,mod-2)%mod
#define ncr(n,r) (((fact[n]*modInv(fact[r]))%mod)*modInv(fact[n-r]))%mod
#define setbits(y) __builtin_popcountll(y)
#define mod 1000000007

ll po(ll a,ll b){
    if(b==0) return 1;
    ll ans=1;
    if(b%2==0){
        ans = po(a,b/2)%mod;
        ans=(ans%mod * ans%mod)%mod;
    }
    else{
        ans = po(a,(b-1)/2)%mod;
        ans = (ans%mod * ans%mod * a)%mod;
    }
    return ans%mod;
}

/////////////////////////////////////////
// Separate string at a particular points
vector <string> tokens; 
void separate_string(string line){
    stringstream check1(line); 
    string lav; 
    while(getline(check1, lav, '*')) 
    { 
        tokens.push_back(lav); 
    } 
}
/////////////////////////////////////////


////////////////////////////////
// All divisors in (sqrt(n));

vector<ll>divisors;
void get_Divisors(ll n){
    for(int i=1;i<=sqrt(n);i++){
        if(n%i==0){
           if(n/i==i){
               divisors.pb(n/i);
           }else{
               divisors.pb(i);
               divisors.pb(n/i);
           }
        }
    }
}

bool sortbysec(const pair<int,int> &a, 
              const pair<int,int> &b) 
{ 
    return (a.second < b.second); 
} 

vector<ll>prim;
void primeFactors(ll n)  
{  
    while (n % 2 == 0)  
    {  
        prim.pb(2); 
        n = n/2;  
    }  
    for (ll i = 3; i <= sqrt(n); i = i + 2)  
    {  
        while (n % i == 0)  
        {  
            prim.pb(i); 
            n = n/i;  
        }  
    }  
    if (n > 2)  
        prim.pb(n);  
}  

////////////////////////////////
////////////////////////////////

// string to int :::::::::::::::::::(stoi());
// string to long long :::::::::::::::::::  (stoll())
// int or long long to string  ::::::::::::::::::: to_string(num);
// gives part before . :::::::::::::::::::  trunc(num);
// rounding numbers :::::::::::::::::::  round(num);
// provides fixed number after decimal places  ::::::::::: cout<<fixed<<setprecision(5)<<p;
// Fermat little theorem pow(a,m-1) congruent to  1(mod m) if a and m are coprime ;
// [a/b] === [a+b-1/b];

///*** Multimap:- ///***
///*** multimap<int,int>mp2  ***///
/// It allows same repeated values as key
///*** mp2.insert(pair <ll, ll> (x, y)) ///***

// nth catalan number 1/(n+1)(2n C n);


//Lower Bound and Upper Bound //
// lower bound of x returns an iterator pointing to the first occurance of element(num) s.t. num>=x else returns v.end();
// Upper bound of x returns an iterator pointing to the first occurance of element(num) s.t. num>x else returns v.end();

//Ordered Set //
//find_by_order(k): It returns to an iterator to the kth element (counting from zero) in the set in O(logn) time.
//order_of_key(k) : It returns to the number of items that are strictly smaller than our item k in O(logn) time.


// Multiset
// multiset<int>st;
// st.insert();
// st.loer_bound();st.upper_bound(),distance(iterato1 , iterator 2);
void solve(){
   ll n,k;
   cin>>n>>k;
   vector<ll>num;
   ll la  = 2*k-n;
   int i=1;
   while(i<la){
       num.pb(i);
       i++;
   }
   
   while(k>=la){
       num.pb(k);
       k--;
   }
   for(auto it:num) cout<<it<<" ";
}


int main(){
    IOS
    test{
        solve(); cout<<"\n";
    }
    // solve();
}