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<iostream>
#include<string>
using namespace std;

int main() {
	int t, n, k; scanf("%d", &t);
	while(t--) {
		scanf("%d %d", &n, &k);
		if(n < (2*k) + 1) printf("-1");
		else if(n == (2*k) + 1) {
			printf("1 ");
			for(int i = 1; i < n; i += 2) {
				printf("%d %d ", i+2, i+1);
			}
		} else {
			int i = 1, peak = 0;
			printf("1 ");
			while(peak != k) {
				printf("%d %d ", i+2, i+1);
				i += 2;
				peak++;
			}
			while(i < n) {
				printf("%d ", i+1);
				i++;
			}
		}
		printf("\n");
	}

	return 0;
}