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:
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC optimize("Ofast")
#include <iostream>
#include <cmath>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <iomanip>
#include <algorithm>
#include <map>
#include <set>
#include <unordered_set>
#include <unordered_map>
#include <vector>
#include <queue>
#include <ctime>
#include <random>
#include <chrono>
#include <stack>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> pll;
typedef pair<ld, ld> pld;
typedef vector<ll> vi;
typedef vector<vi> vvi;

const ll inff = (1e18), mx = 5e6 + 10, MOD = 1e9 + 7, P1 = 61, P2 = 239017, mx2 = 61;
const ld pi = acos(-1), eps = 1e-8;

#define ff first
#define ss second
#define vec vector
#define all(a) a.begin(), a.end()
#define printtime cout << "\nTime: " << 1.0 * clock() / CLOCKS_PER_SEC << " sec\n";
#define rall(a) a.rbegin(), a.rend()
#define pb push_back
#define files(a, b) freopen(a, "r", stdin); freopen(b, "w", stdout);

mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());

/*struct hasher{	
    size_t operator()(const pll &p) const {
		size_t res = p.x * 1e7 + p.y;
		return res;
	}
	hash<ll> llhash;
};*/

ll gcd(ll a, ll b) {
	return b ? gcd(b, a%b) : a;
}

void run() {
	ll n, k;
	cin >> n >> k;
	ll cnti = 0;
	// for (ll i = 1; i <= k; ++i) {
	// 	ll l = k + (k - i);
	// 	if (l < n)
	// 		cnti += n - l;
	// }
	// for (ll i = k + 1; i <= n; ++i) {
	// 	cnti += n - i;
	// }
	// cout << cnti << '\n';
	ll j = 0;
	for (ll i = 1; i <= 2*k - 1 - n; ++i) {
		cout << i << ' ';
		j = i;
	}
	for (ll i = k; i > j; --i)
		cout << i << ' ';
	cout << '\n';
}

signed main() {
	// files("input.txt", "output.txt");
	// files("stair.in", "stair.out");
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
    srand(unsigned(time(0)));
	ll kk = 1;
	cin >> kk;
	while (kk--) {
		run();
		//cout << '\n';
	}
	//system("pause");
	return 0;
}