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: Wrong Answer)

def ascan; gets.split.map(&:to_i);end
def zip(arr)
    arr.each_with_index.to_a.sort.each_with_index.map{|pair,i|[pair[1],i]}.sort.map{|pair|pair[1]+1}
end

N,A,B = ascan

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

if N+1 == A + B
    a = A
    b = B
    ai = 0
    bi = 1000000000
    arr = N.times.map do
        if a > b
            a-=1
            ai+=1
        else
            b-=1
            bi+=1
        end
    end
    puts arr*" "
    exit
end

if A >= B
    arr = []
    ptr = 0
    rem = N - B
    (B).times do |i|
        row = [A-1, rem].min+1
        rem -= row-1
        arr += (row).times.map{|e|e+ptr}.to_a
        ptr -= N
    end
    puts zip(arr)*" "
    exit
end

if A < B
    arr = []
    ptr = 0
    rem = N - A
    (A).times do |i|
        row = [B-1, rem].min+1
        rem -= row-1
        arr += (row).times.map{|e|e+ptr}.to_a.reverse
        ptr += N
    end
    puts zip(arr)*" "
    exit
end

p -1