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:
// #pg199 aka Parth Gulati

#include<bits/stdc++.h>
// #include <ext/pb_ds/assoc_container.hpp>
// using namespace __gnu_pbds;
using namespace std;
#define MOD (int) 1e9+7
#define all(a) a.begin(),a.end()
#define maxpq priority_queue<int>
#define minpq priority_queue<int, vector<int>, greater<int> >
#define fo(i, n) for (int i = 0; i < n; i++)
#define for1(i, n) for (int i = 1; i <= n; i++)
#define forK(i,k,n) for (int i = k; i <= n; i++)
#define loop(i,a,b) for (int i = a; i < b; i++)
#define bloop(i,a,b) for (int i = a ; i>=b;i--)
#define tc(t) int t; cin >> t; while (t--)
#define int long long
#define pb emplace_back
#define prec(n, p) fixed<<setprecision(n)<<p
#define inf (int)(1e18)
#define ini(a, i) memset(a, i, sizeof(a))
#define vi vector<int>
#define fi first
#define se second
#define endl '\n'
#define pii pair<int, int>
#define deb(x)  cout<<#x<<": "<<x<<" "<<endl;
#define vii vector<ii>
#define sz(s) s.size()
#define mInt map<int, int>
#define setbits(n) __builtin_popcount(n)
#define ll long long
#define sl(x) scanf("%lld", &x)
#define pl(x) printf("%lld\n", x)
#define ou cout<<'\n';
#define mp make_pair

//Policy based Data Structure
//typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;
//insert
#define findElement(n) find_by_order(n)
#define findIndex(n) order_of_key(n)

int gcd(int a, int b) { if (a == 0) return b; return gcd(b % a, a);}
int max(int a, int b) {if (a > b) return a; else return b;}
int min(int a, int b) {if (a < b) return a; else return b;}

template<typename... T>
void in(T&... args) {
    ((cin >> args), ...);
}
 
template<typename... T>
void out(T&&... args) {
    ((cout << args << " "), ...);
}
 
template<typename... T>
void outl(T&&... args) {
    ((cout << args << endl), ...);
}

 
int32_t main(){
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    #ifndef ONLINE_JUDGE
		freopen("input1.txt", "r", stdin);
		freopen("output1.txt", "w", stdout);
	#endif
	tc(t){
		int n, k;
		in(n, k);
		int x  = k;
		int y  = 2*k-n;
		vector<int> ans;
		vector<int> ans2;

		for(int i = 1;i<y;i++){
			ans.pb(i);
		}
		int z  = k;
		for(int i = k;i>=y;i--){
			ans.pb(i);
		}
		fo(i, k)out(ans[i]);
		ou;
	}
    
}