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;
typedef long long ll;
#define endl "\n"
#define space " "
#define ll long long int
#define mod 1000000007
#define pi 3.1415926535897932384626433
#define inf 1000000000000000001;
#define fast ios_base::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL);
#define all(c) c.begin(),c.end()
#define mp(x,y) make_pair(x,y) 
#define mem(a,val) memset(a,val,sizeof(a))
#define pb push_back
#define f first
#define s second
#define fl(i,x,n)  for(ll i=x; i<n; i++)
#define fr(i,x,n)  for(ll i=n-1; i>=x; i--)
void solve()
{
	ll n,k; cin>>n>>k;
	ll d=n-k;
	std::vector<ll> v(k);
	fl(i,0,k) v[i]=i+1;
	ll p=k-1-d;
	reverse(v.begin()+p, v.end());
	fl(i,0,k) cout<<v[i]<<space;
	cout<<endl;
}

signed main()
{

fast;
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif

ll T=1;
cin>>T;
while(T--)
{
	solve();
}
return 0;
}