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 all(n)      for(int i=0; i<int(n); i++)
#define rep(i,a,b)  for(int i=a; i<=(int)b; i++)
#define rrep(i,a,b) for(int i=a; i>=(int)b; i--)
#define pb          push_back
#define pii         pair<int, int>
#define pll         pair<ll, ll>
#define vi          vector<int>
#define vvi         vector<vi>
#define vll         vector<ll>
#define F           first
#define S           second
#define             endl '\n'
#define FIO         ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
 
void test_case() {
    int n, k;
    cin>>n>>k;
 
    int i;
    vector<int> v;
    for(i=k; i>=k-(n-k); i--) v.pb(i);
 
    vector<int> vv;
    for(int j=1; j<=i; j++) cout<<j<<" ";
 
    // reverse(v.begin(), v.end());
    for(int x: v) cout<<x<<" ";
    cout<<'\n';
}
 
int main(){
    FIO;
 
    int t=1;
    cin>>t;
    while(t--) test_case();
}