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:
//Code by Ritik Agarwal
#include<bits/stdc++.h>
using namespace std;
#define sz(x) (int)(x).size() 
#define int long long int
#define loop(i,a,b) for(int i=a;i<b;i++)
#define scan(arr,n) for (int i = 0; i < n; ++i) cin >> arr[i]
#define vi vector<int>
#define si set<int>
#define pii pair <int, int>
#define sii set<pii>
#define vii vector<pii>
#define mii map <int, int>
#define pb push_back
#define ff first
#define ss second
#define all(aa) aa.begin(), aa.end()
#define rall(a) a.rbegin() , a.rend()
#define read(a,b) int a,b;  cin>>a>>b
#define readt(a,b,c) int a,b,c; cin>>a>>b>>c
#define readf(a,b,c,d) int a,b,c,d; cin>>a>>b>>c>>d;
#define print(v) for(auto x:v) cout<<x<<" ";cout<<endl
#define printPair(res) for(pair<int,int>& p:res) cout<<p.first<<" "<<p.second<<endl;
#define faster ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
//*****  Constants  *****
const int mod = 1000000007; /* 1e9 + 7*/
const int MAXN = 1500005; /*1e6 +5 */
const int mod1= 998244353;
const int NAX=2E5+5;
int vis[MAXN]={0};
int fac[300001];
void fact(int m){fac[0]=1;for(int i=1;i<=300000;i++) fac[i]=(fac[i-1]%m * i%m)%m;}
/* Iterative Function to calculate (x^y)%p in O(log y)*/
long long power( long long x,  int y, int p){   long long res = 1;  
x = x % p; while (y > 0) { if (y & 1)  res = (res * x) % p;  y = y >> 1; x = (x * x) % p;} return res;}
// Returns n^(-1) mod p 
long long modInverse( long long n, int p) {  return power(n, p - 2, p); }
long long nCr( long long n,int r, int p) { if (r == 0) return 1; if(n<r) return 0;
return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) % p; } 

int binarySearch(vi arr, int l, int r, int x) { 
if (r >= l) { int mid = l + (r - l) / 2; if (arr[mid] == x) return mid; 
if (arr[mid] > x)  return binarySearch(arr, l, mid - 1, x); 
return binarySearch(arr, mid + 1, r, x); } return -1; } 
bool sortGrt(const pair<int,int> &a, 
              const pair<int,int> &b) 
{ 
  
    return (a.second > b.second); 
  
} 
void solve(int test)
{
  read(n,k);
  int req=n-k+1;
  vi v;
  for(int i=1;i<=k-req;i++)
    v.pb(i);
  for(int i=k;i>=k-req+1;i--)
    v.pb(i);

  print(v);
}

int32_t main()
{
      #ifndef ONLINE_JUDGE
          freopen("input.txt", "r", stdin);
          freopen("output.txt", "w", stdout);
      #endif
      faster;
      fact(mod);
      int t=1;
      cin>>t;
      for(int test=1;test<=t;test++)
      {
          solve(test);
      }
}