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:

// Problem: A. Array and Peaks
// Contest: Codeforces - Divide by Zero 2021 and Codeforces Round #714 (Div. 2)
// URL: https://codeforces.com/contest/1513/problem/0
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// Powered by CP Editor (https://github.com/cpeditor/cpeditor)

#include<bits/stdc++.h>
using namespace std;
const int INF = INT_MAX;
const int MOD = 1e9 + 7;
//const int ULTRA = 1e14;
const int PROMAX = 1e6 + 5;
const int PRO = 1e5 + 5;

#define ll long long
#define int ll
#define all(x) x.begin(), x.end()
#define lla(x) x.rbegin(), x.rend()
#define pb push_back
#define ThinkTwice ios_base::sync_with_stdio(false); cin.tie(NULL);
int    dx[] = {-1,  0,  1,  0, -1, -1,  1,  1};
int    dy[] = { 0,  1,  0, -1, -1,  1,  1, -1};
char  dir[] = {'U','R','D','L'};
inline int powmod(int a, int b, int mod) {int res=1; while(b>0) {if(b&1) res=(res*a)%mod; a=(a*a)%mod; b>>=1; } return res;}
inline int power(int a, int b) {int res=1; while(b>0) {if(b&1) res*=a; a*=a; b>>=1;} return res; }
//__________________________________________________________________


void solve() {
    int n, k; cin>>n>>k;
    if(k > (n-1)/2) {
        cout<<-1<<endl;
        return ;
    }
    vector <int> a(n, -1);
    
    int tmp = n;
    int i = 1, k1 = k;
    while(i < n && k) {
        a[i] = tmp--;
        i += 2;
        k -= 1;
    }
    int low = 1;
    for(int i = 0; i<n; i++) {
        if(a[i] == -1) a[i] = low++;
    }
    for(int i = 0; i<n; i++) cout<<a[i]<<" ";
    cout<<endl;
    
}

signed main() {
	ThinkTwice
	int t;
	
	// t = 1;
	cin>>t;

	for(int i = 1; i<=t; i++) {
		// cout<<"Case #"<<i<<": ";
		solve();
	}
}

//__________________________________________________________________
/*
Sample Input:


Sample Output:


*/