/*
 _    _    _______   _    _
| |  / /  |  _____| | |  / /
| | / /   | |       | | / /
| |/ /    | |_____  | |/ /
| |\ \    |  _____| | |\ \
| | \ \   | |       | | \ \
| |  \ \  | |_____  | |  \ \
|_|   \_\ |_______| |_|   \_\

*/
#include <bits/stdc++.h>

using namespace std;

typedef unsigned long long ull;
typedef long long ll;
typedef double ld;
typedef pair <int, int> PII;
typedef pair <ll, ll> PLL;
typedef pair < ll, int > PLI;


#define F first
#define S second
#define pb push_back
#define eb emplace_back
#define right(x) x << 1 | 1
#define left(x) x << 1
#define forn(x, a, b) for (int x = a; x <= b; ++x)
#define for1(x, a, b) for (int x = a; x >= b; --x)
#define mkp make_pair
#define sz(a) (int)a.size()
#define all(a) a.begin(), a.end()
#define y1 kekekek

#define fname ""

const ll ool = 1e18 + 9;
const int oo = 1e9 + 9, base = 1e9 + 7;
const ld eps = 1e-7;
const int N = 2e6 + 6;

int a[N];

void solve() {
    int n, k;
    cin >> n >> k;
    forn(i, 1, k) a[i] = i;
    reverse(a + (k - (n - k)), a + k + 1);

    forn(i, 1, k) {
        cout << a[i] << " ";
    }
    cout << "\n";
}

void clear() {}

int main() {
    ios_base :: sync_with_stdio(0), cin.tie(0), cout.tie(0);

    #ifdef krauch
        freopen("input.txt", "r", stdin);
    #else
        //freopen(fname".in", "r", stdin);
        //freopen(fname".out", "w", stdout);
    #endif

    int T = 1;
    cin >> T;
    while (T--) {
        solve();
        clear();
    }

    return 0;
}

### Prompt

Create a solution in CPP for 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 unsigned long long ull;
typedef long long ll;
typedef double ld;
typedef pair <int, int> PII;
typedef pair <ll, ll> PLL;
typedef pair < ll, int > PLI;


#define F first
#define S second
#define pb push_back
#define eb emplace_back
#define right(x) x << 1 | 1
#define left(x) x << 1
#define forn(x, a, b) for (int x = a; x <= b; ++x)
#define for1(x, a, b) for (int x = a; x >= b; --x)
#define mkp make_pair
#define sz(a) (int)a.size()
#define all(a) a.begin(), a.end()
#define y1 kekekek

#define fname ""

const ll ool = 1e18 + 9;
const int oo = 1e9 + 9, base = 1e9 + 7;
const ld eps = 1e-7;
const int N = 2e6 + 6;

int a[N];

void solve() {
    int n, k;
    cin >> n >> k;
    forn(i, 1, k) a[i] = i;
    reverse(a + (k - (n - k)), a + k + 1);

    forn(i, 1, k) {
        cout << a[i] << " ";
    }
    cout << "\n";
}

void clear() {}

int main() {
    ios_base :: sync_with_stdio(0), cin.tie(0), cout.tie(0);

    #ifdef krauch
        freopen("input.txt", "r", stdin);
    #else
        //freopen(fname".in", "r", stdin);
        //freopen(fname".out", "w", stdout);
    #endif

    int T = 1;
    cin >> T;
    while (T--) {
        solve();
        clear();
    }

    return 0;
}

```