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 int long long
#define all(x) x.begin() , x.end()
#define ld long double
const long long inf = 1e18 + 1LL;
const int mod = 1e9 + 7LL;
void solve();
void ITO();
int32_t main() {
	ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); ITO();
	int t = 1;
	cin >> t;
	while (t--) solve();
	return 0;
}
const int N = (1e6 + 5);
void solve()
{
	int n, k; cin >> n >> k;
	int arr[n];
	int j = 0;
	for (int i = 1; i <= k; i++)
	{
		arr[j++] = i;
	}
	int x = k - 1;
	while (j < n)
	{
		arr[j++] = x--;
	}
	//for (auto x : arr) cout << x << " ";
	vector<int> res;
	for (int i = 1; i < arr[n - 1]; i++) res.push_back(i);
	for (int i = k - 1; i < n; i++) res.push_back(arr[i]);
	for (auto x : res) cout << x << " ";
	cout << "\n";
	return ;
}
void ITO()
{
#ifndef ONLINE_JUDGE
	freopen("inputf.in", "r", stdin);
	freopen("outputf1.in", "w", stdout);
#endif
}
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