Problem: A sequence of n integers is called a permutation if it contains all integers from 1 to n exactly once.

Given two integers n and k, construct a permutation a of numbers from 1 to n which has exactly k peaks. An index i of an array a of size n is said to be a peak if 1 < i < n and a_i \gt a_{i-1} and a_i \gt a_{i+1}. If such permutation is not possible, then print -1.

Input

The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases.

Then t lines follow, each containing two space-separated integers n (1 ≤ n ≤ 100) and k (0 ≤ k ≤ n) — the length of an array and the required number of peaks.

Output

Output t lines. For each test case, if there is no permutation with given length and number of peaks, then print -1. Otherwise print a line containing n space-separated integers which forms a permutation of numbers from 1 to n and contains exactly k peaks. 

If there are multiple answers, print any.

Example

Input


5
1 0
5 2
6 6
2 1
6 1


Output


1 
2 4 1 5 3 
-1
-1
1 3 6 5 4 2

Note

In the second test case of the example, we have array a = [2,4,1,5,3]. Here, indices i=2 and i=4 are the peaks of the array. This is because (a_{2} \gt a_{1} , a_{2} \gt a_{3}) and (a_{4} \gt a_{3}, a_{4} \gt a_{5}). 

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
const int N = 110;
int a[N],n,k;
int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		scanf("%d%d",&n,&k);
		int t=n,p=k;
		for(int i=1;i<=n;++i) a[i]=0;
		for(int i=2;i<=n&&k;i+=2){
			a[i]=t--;
			k--;
		}
		for(int i=1;i<=n;++i)
			if(!a[i]) a[i]=t--;
		int cnt = 0;
		for(int i=2;i<=n-1;++i){
			if(a[i]>a[i-1]&&a[i]>a[i+1]&&i+1<=n&&i-1>=1) cnt++;
		}
		if(cnt!=p) k=10;
		if(k) puts("-1");
		else{
			for(int i=1;i<=n;++i)
				printf("%d ",a[i]);
			puts("");
		}
	}
}