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 int long long
#define ff first
#define ss second
#define endl "\n"
#define all(x) (x).begin(), (x).end()
#define sz(x) (int)(x.size())
int powmod(int a,int l, int md){a%=md; int res=1;while(l){if(l&1)res=res*a%md;l/=2;a=a*a%md;}return res;}
int binpow(int a,int l){int res=1;while(l){if(l&1)res=res*a;l/=2;a=a*a;}return res;}
int invmod(int a, int md){return powmod(a,md-2,md);}
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
typedef vector<int> vi; typedef pair<int, int> ii; typedef vector< ii > vii;
#define pb push_back
int __set(int b, int i) {return b|(1LL<<i);} //set ith bit
int __unset(int b, int i) {return b&(~(1UL<<i));}
int __check(int b, int i) {return b&(1LL<<i);} //returns 0 if ith bit is 0
int mulmod(int a, int b, int md) {return (((a%md)*(b%md))%md+md)%md;}
int addmod(int a, int b, int md) {return ((a%md+b%md)%md+md)%md;}
int submod(int a, int b, int md) {return (((a%md-b%md)%md)+md)%md;}
int divmod(int a, int b, int md) {return mulmod(a, powmod(b, md-2, md), md);} //if md is prime;
const ll inf = 1e9; //very large number
priority_queue<int, vector<int>, greater<int> > pq; //for min priority_queue
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);


signed main(void)
{
	IOS;
	int tt; cin>>tt;
	while(tt--) {
		int n,k; cin>>n>>k;
		vi ans;
		for(int i = 0; i < k; i++)
			ans.pb(i+1);
		// cout<<2*k-n+1<<endl;
		sort(ans.begin()+2*k-n-1,ans.end(),greater<int>());
		for(int i: ans)
			cout<<i<<" ";
		cout<<endl;
	}

	return 0;
}