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;

#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl "\n"
#define ll long long
#define pii pair<int,int>
#define all(x) begin(x), end(x)
#define loop(i,n) for(int i=0; i<n; i++)
#define rep(i,a,b,c) for(int i=a; i<b; i+=c)
#define brep(i,a,b,c) for(int i=a; i>b; i-=c)
#define tc(t) int t; cin>>t; while(t--)
#define sz(v) int((v).size())
#define pb push_back

/*
    Idea:
*/
vector<int> res;
set<int> st;

// bool getPerm (int k, int count, int idx) {
//     if (count < 0) return false;
//     if (count == 0 && idx>=k-1) return true;
//     if (count == 0 || idx>=k-1) return false;

//     for (int i=k; i>=1; i--) {
//         if (st.count(i)) continue;
//         st.insert(i);
//         int n = ;
//         if (getPerm(k, count - , idx+1)) {
//             res[idx] = i;
//             st.erase(i);
//             return true;
//         }
//         st.erase(i);
//     }
// }

int32_t main() {
    IOS;
    int t;
    cin>>t;
    while (t--) {
        int n, k;
        cin>>n>>k;
        vector<int> res(k);
        for (int i=0; i<k; i++)
            res[i] = i+1;
        int diff = k-(n-k);
        if (diff>0) {
            for (int i=diff-1, j=k-1; i<j; i++, j--)
                swap(res[i], res[j]);
        }
        for (auto r : res)
            cout<<r<<" ";
        cout<<"\n";
    }
}