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 (2.7.1) (judge verdict: Accepted)

N, A, B = gets.split.map(&:to_i)

if N < A + B - 1
  puts -1
  exit
end

loop_cnt = Rational(N, A).ceil

if loop_cnt > B
  puts -1
  exit
end

len = (0...B).find { |x|
  lc = Rational(N - x, A).ceil

  lc + x == B
}

ans = []
remain = N

max = N
min = len + 1

tail = [*1...min].reverse

while min <= max
  from = [max - A + 1, min].max
  to = max

  # p [:from, from, :to, to]

  from.upto(to) do |n|
    ans << n
  end

  max = from - 1
end

ans += tail
puts ans.join(' ')