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>
 
#define nl cout<<endl;
#define tb cout<<" ";
#define db cout<<"A";
#define forz(i,n) for(i=0;i<n;i++)
#define forzz(i,n) int i; for(i=0;i<n;i++)
#define pb push_back
#define popb pop_back
#define all(x) x.begin(),x.end()
 
 
#define mp make_pair
#define fi first
#define se second
#define kira ios_base::sync_with_stdio(false); cin.tie(NULL);
#define PI 3.14159265
using namespace std;
 
typedef long long int lli;
typedef pair<int,int> pii;
 
class desc {
    public:
    bool operator() (const lli &a, const lli &b) {
        return a>b;
    }
};

bool pairone(pair<lli,lli> &a, pair<lli,lli> &b) {return (a.fi>b.fi);} 
void show(vector<int> vec) {for(auto x: vec) cout<<x<<" "; nl;}
void showl(vector<lli> vec) {for(auto x: vec) cout<<x<<" "; nl;}

bool divi(string, string);

int main() {
    kira;
    int t; cin>>t;
    while(t--) {
        int n,k;
        cin>>n>>k;
        vector<int> p(k+1);
        for(int i=1; i<2*k-n; i++) p[i]=i;
        int step=k;
        for(int i=2*k-n; i<=k; i++) p[i]=step--;
        for(int i=1; i<=k; i++) cout<<p[i]<<" "; nl;
    }
    return 0;
}