#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> ii;
typedef vector<ii> vii;
#define FOR(i, a, b) for(int i = a; i < b; i++)
#define ROF(i, a, b) for(int i = a; i >= b; i--)
#define ms memset
#define pb push_back
#define F first
#define S second
ll MOD = 1000000007;
 
ll power(ll base, ll n){
	if (n == 0) return 1;
	if (n == 1) return base;
	ll halfn = power(base, n/2);
	if (n % 2 == 0) return (halfn * halfn) % MOD;
	return (((halfn * halfn) % MOD) * base) % MOD;
}
 
ll inverse(ll n){
	return power(n, MOD-2);
}
 
ll add(ll a, ll b){
	return (a+b) % MOD;
}
 
ll mul(ll a, ll b){
	return (a*b) % MOD;
}

int gcd(int a, int b){
	if (a == 0) return b;
	if (a == 1) return 1;
	return gcd(b%a, a);
}
int main(){
	ios::sync_with_stdio(false);
	int TC; cin >> TC;
	while (TC--){
		int n, k; cin >> n >> k;
		vi p(k+1); FOR(i, 1, k+1) p[i] = i;
		int r = k;
		FOR(i, 2*k-n, k+1){
			if (r <= i) break;
			swap(p[i], p[r]);
			r--;
		}
		FOR(i, 1, k+1) cout << p[i] << " ";
		cout << '\n';
	}
	return 0;
}



### Prompt

Please create a solution in CPP to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> ii;
typedef vector<ii> vii;
#define FOR(i, a, b) for(int i = a; i < b; i++)
#define ROF(i, a, b) for(int i = a; i >= b; i--)
#define ms memset
#define pb push_back
#define F first
#define S second
ll MOD = 1000000007;
 
ll power(ll base, ll n){
	if (n == 0) return 1;
	if (n == 1) return base;
	ll halfn = power(base, n/2);
	if (n % 2 == 0) return (halfn * halfn) % MOD;
	return (((halfn * halfn) % MOD) * base) % MOD;
}
 
ll inverse(ll n){
	return power(n, MOD-2);
}
 
ll add(ll a, ll b){
	return (a+b) % MOD;
}
 
ll mul(ll a, ll b){
	return (a*b) % MOD;
}

int gcd(int a, int b){
	if (a == 0) return b;
	if (a == 1) return 1;
	return gcd(b%a, a);
}
int main(){
	ios::sync_with_stdio(false);
	int TC; cin >> TC;
	while (TC--){
		int n, k; cin >> n >> k;
		vi p(k+1); FOR(i, 1, k+1) p[i] = i;
		int r = k;
		FOR(i, 2*k-n, k+1){
			if (r <= i) break;
			swap(p[i], p[r]);
			r--;
		}
		FOR(i, 1, k+1) cout << p[i] << " ";
		cout << '\n';
	}
	return 0;
}



```