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>
#define pf printf
#define sc(x) scanf("%d", &x)
#define scs(x) scanf("%s", x)
#define scl(x) scanf("%lld", &x)
#define mst(a,x) memset(a, x, sizeof(a))
#define rep(i,s,e) for(int i=s; i<e; ++i)
#define dep(i,e,s) for(int i=e; i>=s; --i)
#define INF 0x3f3f3f3f
using namespace std;
typedef  long long ll;
typedef pair<ll,ll> pii;
const long long MOD = 1000000007;
int main()
{
	int t;sc(t);
	while(t--){
		int n,k;sc(n);sc(k);
		if((n % 2 == 1 && k > n / 2) || (n % 2 == 0 && k >= n / 2)){
			pf("-1\n");
			continue;
		}
		int l = 1;
		int r = n;
		rep(i,0,k){
			pf("%d %d ",l,r);
			l++;
			r--;
		}
		if(l == r){
			pf("%d\n",l);
			continue;
		}
		rep(i,l,r)pf("%d ",i);
		pf("%d\n",r);
	} 
}