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 Rust (1.42.0) (judge verdict: Accepted)

use proconio::input;

fn main() {
    input!(n: usize, a: usize, b: usize);
    if n < a + b - 1 || a * b < n {
        println!("-1");
        std::process::exit(0);
    }
    let part: Vec<_> = (0..=a).map(|i| (i * b).min(n - a + i)).collect();
    let mut ans: Vec<_> = (0..n).collect();
    for i in 0..a {
        let l = part[i];
        let r = part[i + 1];
        ans[l..r].reverse();
    }
    println!(
        "{}",
        ans.into_iter()
            .map(|x| (x + 1).to_string())
            .fold("".to_string(), |acc, s| if acc.is_empty() {
                s
            } else {
                acc + " " + s.as_ref()
            })
    );
}