import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int q = in.nextInt();
        for(int a0 = 0; a0 < q; a0++){
            int len = in.nextInt();
            int c = in.nextInt();
            // your code goes here
            if (c == 0) { //bestest case scenario: already sorted!
                for (int i = 1; i <= len; i++) {
                    System.out.print(i);
                    if (i < len) System.out.print(" ");
                    else System.out.println();
                }
            }
            else {
                int max = ((len - 1) * len) / 2;
                if (max < c) System.out.println("-1");
                else if (max == c) { //worst case scenario: sorted in descending order
                    for (int i = len; i > 0; i--) {
                        System.out.print(i);
                        if (i > 1) System.out.print(" ");
                        else System.out.println();
                    }
                }
                else {
                    int min = 0;

                    //if length is even, best case is sum of all smaller odds, and viceversa
                    if (len % 2 == 0) for (int i = 1; i < len; i += 2) min += i;
                    else for (int i = 2; i < len; i += 2) min += i;

                    if (c < min) System.out.println("-1");
                    else if (c == min) { //best case scenario possible for this problem: sorted in descending order in groups of 2
                        for (int i = len; i > 0; i -= 2) {
                            if (i == 1) System.out.println("1");
                            else {
                                System.out.print((i - 1) + " " + i);
                                if (i > 2) System.out.print(" ");
                                else System.out.println();
                            }                            
                        }
                    }
                    else { //the trickiest part is to get permutations in between
                        
                    }
                }
            }
        }
    }
}