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;
typedef long long ll;
#define fo(i,n) for(i=0;i<n;i++)
#define Fo(i,k,n) for(i=k;i<=n;i++)
#define pa pair<ll,ll>
#define vec vector<ll>
#define pb push_back
#define in insert
#define mp make_pair
#define F first
#define S second
//memset(x, 0, sizeof(x))
ll inf = (ll)1e18;
ll mod = 1e9 + 7;
ll max1 = (ll)1e9;
int main() {
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
	freopen("err.txt", "w", stderr);
#endif
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	ll i, j, k, n, m, ct = 0, t, ans = 0;
	cin >> t;
	while (t--) {
		cin >> n >> m;
		vector<ll> vect;
		for (i = 1; i < 2 * m - n; i++) vect.pb(i);
		for (i = m; i >= (2 * m - n); --i) vect.pb(i);
		for (i = 0; i < vect.size(); i++) cout << vect[i] << " ";
		cout << endl;
	}
	return 0;
}