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:
//CODING ME ACCHHA KARNEKA HAI TERKO TO BAS PRACTICE KARTE REH
#include <bits/stdc++.h>
#define pragma
using namespace std;
#define ll long long
#define ld long double
#define pb push_back
#define pf push_front
#define aisort sort(a,a+n)
#define adsort sort(a,a+n,greater<ll>())

typedef pair<ll, ll> pii;
typedef pair<ll, pii> piii;
typedef vector<ll> vi;
typedef vector<string> vs;
typedef vector<pii> vii;
typedef vector<piii> viii;
typedef set<ll> Set;

const ll maxs = 1e5+3;
const ll maxs1 = 1e3+4;
const ll mod = 1e9 + 7;

//MODULO OPEARTIONS
    inline ll add(ll a, ll b) { return (mod + a + b) % mod; }
    inline ll sub(ll a, ll b) { return (a - b + mod) % mod; }
    inline ll mul(ll a, ll b) { return (1ll * a * b) % mod; }

//POWER
    ll fastpow(ll x,ll y)
    {ll res = 1;x = x % mod;while (y > 0){ if (y & 1)res = (res * x) % mod; y = y >> 1 ; x = (x * x) % mod; } return res;}

//MODULAR INVERSE
    ll inv(ll x){ return fastpow(x , mod - 2);}
    
//PRIME
    bool isPrime(ll n) { if (n <= 1)  return false; if (n <= 3)  return true;if (n%2 == 0 || n%3 == 0) return false; for (int i=5; i*i<=n; i=i+6) if (n%i == 0 || n%(i+2) == 0) return false; return true; } 

ll cnt(ll n)
{
    ll t=0;
    while(n>0)
    {
        n/=10;
        t++;
    }
    return t;
}
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ll t;
    cin>>t;
    while(t--)
    {
        ll n,x,i,flag=0,K;
		cin>>n>>K;
		x = (n+1)/2;
		x--;
		if(K>x){
			cout<<"-1\n";
			continue;
		}
		ll c = 1,cc=n;
		for(i=1;i<=n;i++){
			if(i&1){
				cout<<c<<" ";
				c++;
			}else if(K>0){
				cout<<cc<<" ";
				cc--;
				K--;
			}else{
				cout<<c<<" ";
				c++;
			}
		}
		cout<<"\n";
    }
    return 0;
}