Problem:
Score : 700 points

Problem Statement

Determine if there exists a sequence obtained by permuting 1,2,...,N that satisfies the following conditions:

The length of its longest increasing subsequence is A.

The length of its longest decreasing subsequence is B.

If it exists, construct one such sequence.

Notes

A subsequence of a sequence P is a sequence that can be obtained by extracting some of the elements in P without changing the order.

A longest increasing subsequence of a sequence P is a sequence with the maximum length among the subsequences of P that are monotonically increasing.

Similarly, a longest decreasing subsequence of a sequence P is a sequence with the maximum length among the subsequences of P that are monotonically decreasing.

Constraints

1 \leq N,A,B \leq 3\times 10^5

All input values are integers.

Input

Input is given from Standard Input in the following format:

N A B

Output

If there are no sequences that satisfy the conditions, print -1.

Otherwise, print N integers. The i-th integer should be the i-th element of the sequence that you constructed.

Sample Input 1

5 3 2

Sample Output 1

2 4 1 5 3

One longest increasing subsequence of this sequence is {2,4,5}, and one longest decreasing subsequence of it is {4,3}.

Sample Input 2

7 7 1

Sample Output 2

1 2 3 4 5 6 7

Sample Input 3

300000 300000 300000

Sample Output 3

-1

Solution in Ruby (2.3.3) (judge verdict: Accepted)

n,a,b = gets.chomp.split.map(&:to_i)
if n+1 < a+b || a*b < n
  printf("%d\n",-1)
  exit
end

result = []
min = 1
max = n
possible = true

b -= 1
while min <= max
  if max-a+1 <= min
    result.concat((min..max).to_a)
    break
  else
    result.concat(((max-a+1)..max).to_a)
  end
  max -= a
  a -= 1
  
  if max <= min+b-1
    result.concat((min..max).to_a.reverse)
    break
  else
    result.concat((min..(min+b-1)).to_a.reverse)
  end
  min += b
  b -= 1
end

printf("%s\n",result.join(" "))