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:
//JAI SHREE RAM
// PRABHU BRAHMASTRA PRADAN KARE

#include <bits/stdc++.h>
#include <algorithm>
#include <string>
using namespace std;
//#include <ext/pb_ds/assoc_container.hpp> 
//#include <ext/pb_ds/tree_policy.hpp>
//using namespace __gnu_pbds;
//template <typename T>
//using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define vi             vector<ll> 
#define mi             map<ll,ll>
#define pi             pair<ll,ll>
#define si             set<ll>
#define endl            "\n"
#define ins             insert 
#define hell             1000000007
#define PI              3.1415926535897932384
#define ll              long long 
#define f               first
#define s              second
#define pb             push_back
#define pp             pop_back
#define dash()         cout<<"--------"<<endl
#define see(x)         cout<<#x<<" = "<<x<<endl
#define see2(x,y)      cout<<#x<<" = "<<x<<"::"<<#y<<" = "<<y<<endl
#define fr(n)          for(ll i=0;i<n;i++)
#define FAST           ios_base::sync_with_stdio(false); cin.tie(NULL);
#define narak             998244353 
#define decimal(n)  std::cout << std::fixed;std::cout << std::setprecision(n);
#define make make_pair
#define mem0(a)         memset(a,0,sizeof(a))
#define mem1(a)         memset(a,-1,sizeof(a))
#define ALL(v) v.begin(), v.end()
#define SORT(v) sort(ALL(v))  
#define REVERSE(v) reverse(ALL(v))
#define maxc(v) max_element(all(v))
#define minc(v) min_element(all(v))
#define GCD(m,n) __gcd(m,n)
#define LCM(m,n) m*(n/GCD(m,n))
#define sz(a) (ll)a.size()
#define loop(i,a,b) for(ll i=a;i<=b;i++)
#define inf 2000000000000000000
/*
//std::ifstream in("input.txt");
//std::ofstream out("output.txt");
/*

INT_MIN Minimum value for a variable of type int. -2147483647 
INT_MAX Maximum value for a variable of type int.  2147483647
LLONG_MIN Minimum value for a variable of type long long. -9,223,372,036,854,775,807 
LLONG_MAX Maximum value for a variable of type long long.  9,223,372,036,854,775,807


bool comp(ll a,ll b) 
{ return (a < b); } 
ll spf[1000005];
void build()
{
    for(ll i=2;i<1000005;i++)
    {
        if(!spf[i])
        {
            for(ll j=1;j*i<1000005;j++)
            {
            if(!spf[i*j])
              spf[i*j]=i;  
            }
        }
    }
}




bool sortbysec(const pair<ll,ll> &a, const pair<ll,ll> &b) 
{
  return (a.second < b.second); 
} 




const int N=+1e+5+5;
const int LG=20;
vector<ll>g[N];
ll parent[LG][N];
ll timer=0;
ll tin[N],tout[N],level[N];

int __lca(int u,int v)
{
    if(level[u]<level[v])
        swap(u,v);
    int diff=level[u]-level[v];
    for(int i=LG-1;i>=0;i--)
    {
        if((1<<i) & diff){
            u=parent[i][u];
        }
    }
    if(u==v)
        return u;
    for(int i=LG-1;i>=0;i--)
    {
        if(parent[i][u] and parent[i][u]!=parent[i][v])
        {
            u=parent[i][u];
            v=parent[i][v];
        }
    }
    return parent[0][u];
}

void dfs(ll x)
{
   vis[x]=true;
   disc[x]=low[x]=++t;
   for(ll z:v[x])
   {
     if(!vis[z])
     {
      par[z]=x;
      dfs(z);
      low[x]=min(low[x],low[z]);
      if( low[z]>disc[x])
      {
        mp[{x,z}]=1;
        mp[{z,x}]=1;
        //cout<<x<<" "<<z<<endl;
      }
     }
     else if(z!=par[x])
     {
        low[x]=min(low[x],disc[z]);
     }
   }

   return ;
}
void dfs(ll v,ll par,ll lvl)
{
    arr[v]=cnt;
    tin[v]=++timer;
    parent[0][v]=par;
    level[v]=lvl;
    
    for(auto &u:g[v])
    {
        if(u==par)
            continue;
        dfs(u,v,lvl+1);
    }
    tout[v]=timer;
}
bool is_ancestor(int u,int v)
{
    return (tin[u]<=tin[v] and tout[u]>=tout[v]);
}

ll kthancestor(ll v,ll p)
{
    ll curr=v;
    for(ll i=20;i>=0;i--)
    {
        if( 1<<i & p)
            curr=parent[i][curr];
    }
    return curr;
}







ll n,m;
ll k;

const int K=+1e+5+5;
set<ll>v[K];
vector<ll>tree[K];
bool vis[K];
ll disc[K];
ll low[K];
ll t=0;
const int LG=20;
vector<ll>g[K];
ll parent[LG][K];
ll timer=0;
ll tin[K],tout[K],level[K];
map<pair<ll,ll>,ll>mp;
ll arr[K];
ll par[K];



void fun(ll x,ll cnt)
{
    vis[x]=1;
    //tree[cnt].pb(x);
    arr[x]=cnt;
    for(ll y:v[x])
    {
        if(!vis[y])
            fun(y,cnt);
    }
}





void dfs3(ll v,ll par,ll lvl)
{
    //arr[v]=cnt;
    tin[v]=++timer;
    parent[0][v]=par;
    level[v]=lvl;

    for(auto &u:tree[v])
    {
        if(u==par)
            continue;
        dfs3(u,v,lvl+1);
    }
    tout[v]=timer;
}
bool is_ancestor(int u,int v)
{
    return (tin[u]<=tin[v] and tout[u]>=tout[v]);
}
void precompute()
{
    for(int i=1;i<LG;i++)
    {
        for(int j=1;j<=n;j++)
        {
            if(parent[i-1][j])
            {
                parent[i][j]=parent[i-1][parent[i-1][j]];
            }
        }
    }
}

ll distance(ll u,ll v)
{
    return level[u]+level[v]-2*level[__lca(u,v)];
}

void bfs(ll src,ll dest,ll ans )
{
    vis[src]=1;
    if(src==dest)
    {
        cout<<ans<<endl;
        return ;
    }
    for(ll x:v[src])
    {
        if(!vis[x])
            bfs(x,dest,ans+mp[{x,src}]);
    }
    return ;
}





getline(cin, str);  // to get a line as input
getline(cin, str);








vi fac(200005);

ll power(ll x,ll y,ll p)
{
    ll res = 1; // Initialize result
    x = x % p; // Update x if it is more than or equal to p
    while (y > 0) 
    {
        // If y is odd, multiply x with result
        if (y & 1)
            res = (res * x) % p;
        // y must be even now
        y = y >> 1; // y = y/2
        x = (x * x) % p;
    }
    return res;
}
ll modInverse(ll n,ll p)
{
    return power(n, p - 2, p);
}
ll ncr(ll n,ll k,ll mod) 
{
    ll nom = fac[n];
    ll den = (fac[n-k] * fac[k]) % mod;
    return (nom * power(den, mod - 2,mod)) % mod;
}


*/


//RNG based on mersenne_twister 
//mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());





void solve(ll nn)
{   
    ll n,k;
    cin>>n>>k;
    if(k>(n-1)/2)
    {
        cout<<-1<<endl;
        return ;
    }
    else
    {
        ll ans[n+1];
        mem1(ans);
        ll j=1;
        for(ll i=1;i<=k+1;i++)
        {
            ans[j]=i;
            j+=2;
        }
        ll x=k+2;
        for(ll i=1;i<=n;i++)
        {
            if(ans[i]==-1)
            {
                ans[i]=x;
                x++;
            }
            else
            {
                
            }
            
        }
        for(ll i=1;i<=n;i++)
        cout<<ans[i]<<" ";
        cout<<endl;
    }
}
  
int main() 
{ 
   //decimal(15);
   FAST;
   //build();
   ll t;
   t=1;
   cin>>t;
   for(ll i=1;i<=t;i++)
   solve(i);
   return 0;
}