Absolute Permutation

problem

We define  to be a permutation of the first  natural numbers in the range . Let  denote the position of  in permutation  (please use -based indexing).

 is considered to be an absolute permutation if  holds true for every .

Given  and , print the lexicographically smallest absolute permutation, ; if no absolute permutation exists, print -1.

Input Format

The first line of input contains a single integer, , denoting the number of test cases. 
Each of the  subsequent lines contains  space-separated integers describing the respective  and  values for a test case.

Constraints

Output Format

On a new line for each test case, print the lexicographically smallest absolute permutation; if no absolute permutation exists, print -1.

Sample Input

3
2 1
3 0
3 2
Sample Output

2 1
1 2 3
-1

solution

#!/bin/ruby

t = gets.strip.to_i
for a0 in (0..t-1)
  n,k = gets.strip.split(' ')
  n = n.to_i
  k = k.to_i
    
  
  if (n % 2 != 0 and k != 0)
    puts "-1"
    
  elsif k == 0
    1.upto(n).each {|i|
      print i
      print " "
    }
    puts
    
  else
    ar = []
    c = 0
    1.upto(n).each {|i|
      if !ar[i]
        c += 2
        ar[i] = i + k
        ar[i + k] = i      
      end
    }
  
    if c != n
      puts "-1"
    else
      puts ar[1..-1].join(" ")    
    end
  end
end