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 ll long long
#define pii pair<int, int>
#define pll pair<long long, long long>
#define pb push_back
#define ff first
#define ss second
#define YES cout << "YES\n"
#define NO cout << "NO\n"
#define nn "\n"
#define sci(x) scanf("%d", &x)
#define LL_INF (1LL << 62)
#define INF (1 << 30)
#define SetBit(x, k) (x |= (1LL << k))
#define ClearBit(x, k) (x &= ~(1LL << k))
#define CheckBit(x, k) (x & (1LL << k))
#define mod 1000000007
#define N 100005


int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int t;
    cin >> t;
    while(t--){
        int n, k;
        cin >> n >> k;
        vector<int> v1, v2;
        for(int i = 0; i < n-k+1; i++){
            v1.pb(k-i);
        }

        // if(v1.empty()) v1.pb(k);

        for(int i = 1; i < v1.back(); i++){
            v2.pb(i);
        }

        for(int x: v2)
            cout << x << " ";
        for(int x: v1)
            cout << x << " ";

        cout << nn;


    }
    
    return 0;
}