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 mp make_pair
#define pb push_back
#define SZ(x) (int)(x.size())
#define all(x) x.begin(),x.end()
using namespace std;
typedef long long ll;
typedef pair<int,int>pii;
const int maxn = 1e6 + 6;
const int  inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int N = 2e3 + 3;
ll qpow(ll x,ll y) { ll ans=1;x%=mod;  while(y) { if(y&1) ans=ans*x%mod; x=x*x%mod;  y>>=1; } return ans; }

int _;
int n,k;
int a[maxn];
int v[maxn];
void so() {
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++) a[i]=i,v[i]=0;
    if(k>(n-1)/2) {
        puts("-1");
        return ;
    }
    vector<int>b(n+1,0);
    for(int i=1,l=(n+1)/2+1;i<=n;i++) {
        if(v[i]) continue;
        printf("%d ",i);v[i]=1;
        if(k>0) {
            printf("%d ",l);v[l]=1;k--;l++;
        }
    }
    printf("\n");
}
int main() {
    for(scanf("%d",&_);_;_--) { 
        so();
    }
    return 0;
}