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>


//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
//using namespace __gnu_pbds;

//typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;

#define typeof(x) __typeof__(x)
#define bit(x,i) (x&(1<<i))  //select the bit of position i of x
#define lowbit(x) ((x)&((x)^((x)-1))) //get the lowest bit of x
#define hBit(msb,n) asm("bsrl %1,%0" : "=r"(msb) : "r"(n)) //get the highest bit of x, maybe the fastest
#define max(a,b) (a<b?b:a)
//#define abs(x) (x<0?(-x):x) // big bug here if "-x" is not surrounded by "()"
#define IN(i,l,r) (l<i&&i<r) //the next for are for checking bound
#define LINR(i,l,r) (l<=i&&i<=r)
#define LIN(i,l,r) (l<=i&&i<r)
#define INR(i,l,r) (l<i&&i<=r)
#define F(i,L,R) for (int i = L; i < R; i++) //next four are for "for loops"
#define FIN(i,L,R) for (int i = L; i <= R; i++)
#define FF(i,L,R) for (int i = L; i > R; i--)
#define FFIN(i,L,R) for (int i = L; i >= R; i--)
#define getI(a) scanf("%d", &a) //next three are handy ways to get ints, it's also force you to use '&' sign
#define getII(a,b) scanf("%d%d", &a, &b)
#define getIII(a,b,c) scanf("%d%d%d", &a, &b, &c)

#define getLL(a) scanf("%lld", &a) 
#define get2LL(a,b) scanf("%lld %lld", &a,&b)
#define g_s(n) int (n); scanf("%d",&(n)) //handy if the input is right after the definition of a variable
#define g_s2(n,m) int (n),(m); scanf("%d %d",&(n),&(m))
#define g_s3(n,m,k) int (n),(m),(k); scanf("%d %d %d",&(n),&(m),&(k))
#define TESTS g_s(testow);while(testow--) //for multilple cases problems
#define whileZ int T; getI(T); while(T--) // the same as above
#define getS(x) scanf("%s", x) //get a char* string
#define MEM(a,x) memset(a,x,sizeof(a)) //set elements of array to some value
#define char2Int(c) (c-'0')
#define lastEle(vec) vec[vec.size()-1]
#define SZ(x) ((int)((x).size()))
#define REMAX(a,b) (a)=max((a),(b)) // set a to the maximum of a and b
#define REMIN(a,b) (a)=min((a),(b));
#define FOREACH(i,t) for (typeof(t.begin()) i=t.begin(); i!=t.end(); i++) // traverse an STL data structure
#define ALL(c) (c).begin(),(c).end() //handy for function like "sort()"
#define PRESENT(c,x) ((c).find(x) != (c).end())
#define CPRESENT(c,x) (find(ALL(c),x) != (c).end())
#define ll long long //data types used often, but you don't want to type them time by time
#define ull unsigned long long
#define ui unsigned int
#define us unsigned short
#define IOS ios_base::sync_with_stdio(0); //to synchronize the input of cin and scanf
#define INF 1001001001
#define PI 3.1415926535897932384626
//for map, pair
#define mp make_pair
#define fi first
#define se second
// for debug
inline void pisz(int n) { printf("%d\n",n); }
#define DBG(vari) cerr<<#vari<<" = "<<(vari)<<endl;
#define printA(a,L,R) FIN(jk,L,R) cout << a[jk] << (jk==R?'\n':' ')
#define printV(a) printA(a,0,a.size()-1)

#define sf scanf
#define pf printf
//for vectors
#define pb push_back
typedef int elem_t;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<string> vstr;
typedef vector<set<int> > vset;
typedef vector<vi> vvi;
typedef vector<vll> vvll;
typedef vector<bool> vb;
typedef vector<vector<bool> > vvb;
typedef pair<int,int> ii;
typedef vector<ii> vii;
// directions
const int dr[4] = {-1,0,1,0};
const int dc[4] = {0,-1,0,1};
const int drr[8] = {-1,-1,0,1,1,1,0,-1};
const int dcc[8] = {0,-1,-1,-1,0,1,1,1};
template<typename T,typename TT> ostream& operator<<(ostream &s,pair<T,TT> t) {return s<<"("<<t.first<<","<<t.second<<")";}
template<typename T> ostream& operator<<(ostream &s,vector<T> t){F(i,0,SZ(t))s<<t[i]<<" ";return s; }

//find_by_order(k): It returns to an iterator to the kth element (counting from zero)
//order_of_key(k) : It returns to the number of items that are strictly smaller than our item k

class UnionFind{
    private: vi rank,p;
    public: 
        UnionFind(int n){
            F(i,0,n)p.pb(i);
            rank.assign(n,0);
        }

        int findSet(int s){
            return (p[s]==s) ? s : (p[s]=findSet(p[s]));
        }

        bool isSameSet(int i,int j){
            return findSet(i)==findSet(j);
        }

        void unionSet(int i,int j){
            if(isSameSet(i,j))return;

            int x=findSet(i),y=findSet(j);
            if(rank[x]>rank[y])p[y]=x;
            else {
                p[x]=y;
                if(rank[x]==rank[y])rank[y]++;
            }
            
        }
};

bool fcomp (double A, double B) 
{
	double EPSILON=numeric_limits<double>::epsilon();
    double diff = A - B;
    return fabs(diff)<EPSILON;
}

double x_y(int x1,int y1,int x2,int y2){
    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}



int main(){
    //ios::sync_with_stdio(false);
    //cin.tie(nullptr);
	//freopen ("out.txt","w",stdout);
	TESTS{
		g_s2(n,k);
		if(k>(n-1)/2){
			cout<<"-1"<<endl;
			continue;
		}
		
		int f[n+1];
		MEM(f,0);
		
		int cur=1,v=n;
		vi ans(n,-1);
		//MEM(ans,-1);
		
		while(k--){
			ans[cur]=v;
			f[v]=1;
			cur += 2;
			v--;
		}
		
		F(i,0,n){
			if(ans[i]==-1){
				F(j,1,n+1){
					if(f[j])continue;
					f[j]=1;
					ans[i]=j;
					break;
				}
			}
		}
		
		printV(ans);
	}
		

    return 0;
}