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;

using ll = long long;
using ull = unsigned long long;
#define v(t) vector<t>
#define p(t) pair<t, t>
#define p2(t, s) pair<t, s>
#define vp(t) v(p(t))

#define rep(i, n) for (int i=0,i##_len=((int)(n)); i<i##_len; ++i)
#define rep2(i, a, n) for (int i=((int)(a)),i##_len=((int)(n)); i<=i##_len; ++i)
#define repr(i, n) for (int i=((int)(n)-1); i>=0; --i)
#define rep2r(i, a, n) for (int i=((int)(n)),i##_len=((int)(a)); i>=i##_len; --i)

#define repi(itr, c) for (__typeof((c).begin()) itr=(c).begin(); itr!=(c).end(); ++itr)
#define repir(itr, c) for (__typeof((c).rbegin()) itr=(c).rbegin(); itr!=(c).rend(); ++itr)

#define sz(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()

#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define RSORT(x) sort(rall(x));
#define pb push_back
#define eb emplace_back

#define INF (1e9)
#define LINF (1e18)
#define PI (acos(-1))
#define EPS (1e-7)
#define DEPS (1e-10)

int main(){
    int t;
    cin >> t;
    rep(i1, t) {
        int n, k;
        cin >> n >> k;
        int d = n - k + 1;
        v(int) ans(k);
        int l = k-d, r = k-1;
        int num = k;
        rep2(i, l, r) {
            ans[i] = num;
            --num;
        }
        rep2(i, 0, l-1) ans[i] = i+1;
        rep(i, k) printf("%d%c", ans[i], (i<k-1?' ':'\n'));
    }
    return 0;
}