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

using namespace std;

#define int long long
#define ll long long
#define double long double
#define pii pair<int, int>
#define vec vector
#define u_map unordered_map
#define u_set unordered_set
#define str string
#define all(el) el.begin(), el.end()
#define allr(el) el.rbegin(), el.rend()
#define sz(el) (int)el.size()
#define f first
#define s second
#define x first
#define y second

#define mn first
#define mx second

#define pb push_back
#define in insert
#define mp make_pair
#define endl "\n"
#define forn(i, n) for (int i = 0; i < n; ++i)

template<typename A, typename B> bool mini(A& x, const B& y) {
    if (y < x) {
        x = y;
        return true;
    }
    return false;
}

template<typename A, typename B> bool maxi(A& x, const B& y) {
    if (y > x) {
        x = y;
        return true;
    }
    return false;
}

int gcd(int a, int b) {
    if (b == 0) {
        return a;
    }
    return gcd(b, a % b);
}

int get_lcm(int a, int b) {
    return a * b / gcd(a, b);
}

int check(str &s, str &a) {
    int ok = 1;
    for (int i = 0; i < sz(s); i++) {
        ok &= (s[i] == a[i % sz(a)]);
    }
    return ok;
}

int get_sm(int x) {
    return (0 + x) * x / 2;
}

void smart() {
    int n, k;
    cin >> n >> k;
//    (n - k) + (n - k - 1) + (n - k - 2) + ... (n - k - (n - k))
//    (n - k) * (n - k) - get(1...(n - k))

    int sz2 = (n - k);
    int sz1 = n - sz2 - (sz2 + 1);

    vec<int> ans(k);

    for (int i = 0; i < sz1; i++) {
        ans[i] = i;
    }
    for (int i = sz1; i < k; i++) {
        ans[i] = (k - 1 - (i - sz1));
    }

    for (auto now : ans) {
        cout << now + 1 << ' ';
    }
    cout << endl;

}

signed main() {

    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

//        freopen("/Users/borisbarkov/Desktop/main_clion/input.txt", "r", stdin);

    int t;
    cin >> t;
    while (t--) {
        smart();
    }

    return 0;
}