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 For(i,a,b) for(int i=a,i##_end=b;i<i##_end;++i)
using ll=long long;
using pii=pair<int,int>;
#define fi first
#define se second
#define mp make_pair
#define pb push_back
const int mo=998244353;
int Pow(int a,int b,int p=mo){int r=1;for(;b;b>>=1,a=1ll*a*a%p)if(b&1)r=1ll*r*a%p;return r;}
//template<typename T>
void inc(int &x,int y){x=(x+y>=mo?x+y-mo:x+y);}
template<typename T>T gcd(T a,T b){while(b)a%=b,swap(a,b);return a;}
template<typename T>int upmn(T &x,T y){return x>y?x=y,1:0;}
mt19937 rnd(time(0));
ll sqr(int x){return 1ll*x*x;}
const int N=2e5+11,M=(N<<2)+11;

int n,k;
int p[N];
int main(){
#ifdef SGR
	freopen("a.txt","r",stdin);
#endif
	ios::sync_with_stdio(0);
	int T;
	cin>>T;
	while(T--){
		cin>>n>>k;
		For(i,1,k+1)p[i]=i;
		reverse(p+2*k-n,p+k+1);
	//	cerr<<2*k-n-1<<endl;
		For(i,1,k+1){
			cout<<p[i]<<" \n"[i==k];
		}
	}
}