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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <algorithm>
#include <fstream>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <set>
#include <stdio.h>
#include <string>
#include <utility>
#include <vector>

using namespace std;

string solve(int n, int m, int a, int b, bool mode) {
  if (n == 1) {
    if (a > 0 && b > 0) {
      if (mode)
        return to_string(m);
      else
        return "1";
    } else {
      return "-1";
    }
  }
  if ((b < n && a == 1) || (a < n && b == 1)) {
    return "-1";
  }
  if (a + b > n + 1)
    return "-1";
  vector<int> vl, vr;
  for (int i = 0; i < a; i++) {
    vl.push_back(n - a + i + 1);
  }
  for (int i = 0; i < b - 1; i++) {
    // [n - a - b, n - a - 1]
    vr.push_back(n - a + 2 - b + i);
  }
  if (mode) {
    for (int i = 0; i < vl.size(); i++)
      vl[i] = m - vl[i] + 1;
    for (int i = 0; i < vr.size(); i++)
      vr[i] = m - vr[i] + 1;
  }
  if (n - a - b + 1 == 0) {
    string ret = "";
    ret += to_string(vl[0]);
    for (int i = 1; i < vl.size(); i++)
      ret += " " + to_string(vl[i]);
    for (int i = 0; i < vr.size(); i++)
      ret += " " + to_string(vr[i]);
    return ret;
  }
  string middle = solve(n - a - b + 1, n, a - 1, b - 1, mode);
  if (middle == "-1")
    return "-1";
  string ret = "";
  ret += to_string(vl[0]);
  for (int i = 1; i < vl.size(); i++) {
    ret += " " + to_string(vl[i]);
  }
  ret += " " + middle;
  for (int i = 0; i < vr.size(); i++)
    ret += " " + to_string(vr[i]);
  return ret;
}

int main() {
  int n, a, b;
  cin >> n >> a >> b;
  if (a > b)
    cout << solve(n, n, a, b, false) << endl;
  else
    cout << solve(n, n, b, a, true) << endl;
  return 0;
}