#include <bits/stdc++.h>
#define code                     \
    ios::sync_with_stdio(false); \
    cin.tie(0);                  \
    cout.tie(0);
#define int long long
#define endl '\n'
#define MM(a, b) memset(a, b, sizeof(a))
#define all(v) v.begin(), v.end()
using namespace std;
typedef long long ll;
const ll MOD = 1e9 + 7;

#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char *name, Arg1 &&arg1)
{
    cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char *names, Arg1 &&arg1, Args &&... args)
{
    const char *comma = strchr(names + 1, ',');
    cout.write(names, comma - names) << " : " << arg1 << " | ";
    __f(comma + 1, args...);
}

int32_t main()
{
    code;
    int t;
    cin >> t;
    while (t--)
    {
        int n, k;
        cin >> n >> k;
        vector<int> p(k);
        for (int i = 0; i < k; i++)
        {
            p[i] = i + 1;
        }
        n = n - k + 1;
        reverse(p.begin() + k - n, p.end());
        for (auto i : p)
            cout << i << " ";
        cout << endl;
    }
}

### Prompt

Generate a CPP solution 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>
#define code                     \
    ios::sync_with_stdio(false); \
    cin.tie(0);                  \
    cout.tie(0);
#define int long long
#define endl '\n'
#define MM(a, b) memset(a, b, sizeof(a))
#define all(v) v.begin(), v.end()
using namespace std;
typedef long long ll;
const ll MOD = 1e9 + 7;

#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char *name, Arg1 &&arg1)
{
    cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char *names, Arg1 &&arg1, Args &&... args)
{
    const char *comma = strchr(names + 1, ',');
    cout.write(names, comma - names) << " : " << arg1 << " | ";
    __f(comma + 1, args...);
}

int32_t main()
{
    code;
    int t;
    cin >> t;
    while (t--)
    {
        int n, k;
        cin >> n >> k;
        vector<int> p(k);
        for (int i = 0; i < k; i++)
        {
            p[i] = i + 1;
        }
        n = n - k + 1;
        reverse(p.begin() + k - n, p.end());
        for (auto i : p)
            cout << i << " ";
        cout << endl;
    }
}

```