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>
using namespace std;
#define ll long long
#define f(i,x,n) for(int i=x;i<=n;i++)
#define all(c) c.begin(),c.end()
#define cin(a,n) for(int i=0;i<n;i++){cin>>a[i];}
#define int int64_t
#define pi pair<int,int>
#define countbits __builtin_popcount
const int Nmax=1e9+7;
const int mod=1000000007;
vector<int>prime;


int is_power_of_two(int a){
    return a && (!(a&(a-1)));
}


int isperfectsquare(int a){
    double b=sqrt(a);
    return (b-ceil(b)==0);
}

/* POWER FUNCTION*/


int pow(int a,int b,int m){
    a=a%m;
    int res=1;
    while(b>0){
        if(b&1){
            res=res * a%m;
        }
        a=a * a %m;
        b=b/2;
    }
return res;
}
/* SIEVE OF ERATOSTHENES FOR PRIME */
void seive(int n){
    bool isprime[n];
    memset(isprime,true,sizeof(isprime));
    prime.push_back(2);
    for(int i=3;i<=sqrt(n);i+=2){
        if(isprime[i]){
            for(int j=i*i;j<=n;j+=i){
                isprime[j]=false;
            }
        }
    }
    for(int i=3;i<=n;i+=2){
        if(isprime[i]){
            prime.push_back(i);
        }
    }
}


/*YOU CAN SOLVE IT*/

void solve(){
    int n,k;
    cin>>n>>k;
    if(k>(n-1)/2){
        cout<<"-1\n";
        return;
    }
    if(k==0){
        for(int i=1;i<=n;i++){
            cout<<i<<" ";
        }
        cout<<"\n";
        return;
    }
    if(k>=1){
        cout<<"1 3 2 ";
        k--;
    }
    int p=2;
    for(int i=1;i<=k;i++){
        cout<<p+3<<" "<<p+2<<" ";
        p=p+2;
    }
    p+=2;
    for(int i=p;i<=n;i++){
        cout<<i<<" ";
    }
    cout<<"\n";
}


int32_t main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin>>t;
    while(t--){
        solve();
    }
    return 0;
}