#include<iostream>
#include<cstdio>
#include<algorithm>
namespace do_while_true {
	#define ld double
	#define ll long long
	#define re register
	#define pb push_back
	#define fir first
	#define sec second
	#define pp std::pair
	#define mp std::make_pair
	const ll mod = 1000000007;
	template <typename T>
	inline T Max(re T x, re T y) { return x > y ? x : y; }
	template <typename T>
	inline T Min(re T x, re T y) { return x < y ? x : y; }
	template <typename T>
	inline T Abs(re T x) {	return x < 0 ? -x : x; }
	template <typename T>
	inline void read(re T& r) {
		r = 0; bool w = 0; char ch = getchar();
		while(ch < '0' || ch > '9') w = ch == '-' ? 1 : 0, ch = getchar();
		while(ch >= '0' && ch <= '9') r = r * 10 + (ch ^ 48), ch = getchar();
		r = w ? -r : r;
	}
	template <typename T>
	inline T qpow(re T x, re T y) {
		re T sumq = 1; x %= mod;
		while(y) {
			if(y&1) sumq = sumq * x % mod;
			x = x * x % mod;
			y >>= 1;
		}
		return sumq;
	}
}
using namespace do_while_true;
const int N = 100010;
int n, k, a[N];

void solve() {
	read(n); read(k); int s = n-k+1, down = k-s+1;
	for(int i = 1; i < down; ++i) printf("%d ", i);
	for(int i = k; i >= down; --i) printf("%d ", i);
	puts("");
}

signed main() {
	#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
	#endif
//	freopen("out1.txt", "w", stdout);
	int T = 1;
	read(T);
	while(T--)
		solve();
	fclose(stdin);	
	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<iostream>
#include<cstdio>
#include<algorithm>
namespace do_while_true {
	#define ld double
	#define ll long long
	#define re register
	#define pb push_back
	#define fir first
	#define sec second
	#define pp std::pair
	#define mp std::make_pair
	const ll mod = 1000000007;
	template <typename T>
	inline T Max(re T x, re T y) { return x > y ? x : y; }
	template <typename T>
	inline T Min(re T x, re T y) { return x < y ? x : y; }
	template <typename T>
	inline T Abs(re T x) {	return x < 0 ? -x : x; }
	template <typename T>
	inline void read(re T& r) {
		r = 0; bool w = 0; char ch = getchar();
		while(ch < '0' || ch > '9') w = ch == '-' ? 1 : 0, ch = getchar();
		while(ch >= '0' && ch <= '9') r = r * 10 + (ch ^ 48), ch = getchar();
		r = w ? -r : r;
	}
	template <typename T>
	inline T qpow(re T x, re T y) {
		re T sumq = 1; x %= mod;
		while(y) {
			if(y&1) sumq = sumq * x % mod;
			x = x * x % mod;
			y >>= 1;
		}
		return sumq;
	}
}
using namespace do_while_true;
const int N = 100010;
int n, k, a[N];

void solve() {
	read(n); read(k); int s = n-k+1, down = k-s+1;
	for(int i = 1; i < down; ++i) printf("%d ", i);
	for(int i = k; i >= down; --i) printf("%d ", i);
	puts("");
}

signed main() {
	#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
	#endif
//	freopen("out1.txt", "w", stdout);
	int T = 1;
	read(T);
	while(T--)
		solve();
	fclose(stdin);	
	return 0;
}

```