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>
#define ll long long
#define _n 2020
#define C continue
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define gc getchar
#define pc putchar
#define kg putchar(' ')
#define hh putchar('\n')
#define mem(a,b) memset(a,b,sizeof(a))
#define _for(i,a,b) for( int i=(a); i<=(b); i++)
#define _rep(i,a,b) for( int i=(a); i>=(b); i--)
//#define int __int128
using namespace std;
inline int rd(){
    int x=0,f=1;char ch=gc();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=gc();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=gc();}
    return x*f;
}
char chhh[200];
inline void rt(int x){
    int cnt=0,tmp=x>0?x:-x;
    if (!x) {pc('0');return;}
    if(x<0) pc('-');
    while(tmp>0) {chhh[cnt++]=tmp%10+'0';tmp/=10;}
    while(cnt>0) pc(chhh[--cnt]);
}
//signed main(){
int main(){
	int t;
	t=rd();
	while(t--){
		int n=rd(),k=rd();
		int b=2*k-n;
		for(int i=1;i<b;i++){
			printf("%d ",i);
		} 
		for(int i=k;i>=b;i--){
			printf("%d ",i);
		}
		puts("");
	}
    return 0;
}