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 <bits/stdc++.h>

using namespace std;

typedef long long ll;
const ll INF = 1LL<<60;

ll gcd(ll a, ll b) {
  if(b == 0) return a;
  return gcd(b, a % b);
}

template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }

ll modpow(ll a, ll n, ll mod) {
  ll res = 1;
  while (n > 0) {
    if (n & 1) res = res * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return res;
}

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);

  int N, A, B;
  cin >> N >> A >> B;

  bool is_swaped = false;

  if(A < B){
    swap(A, B);
    is_swaped = true;
  }

  if( B > (N-A)+1){
    cout << -1 << endl;
  }else if(B == 1){
    if(A == N){
      if(is_swaped){
        for(int i = N; i >= 1; i--){
          cout << i << " ";
        }
        cout << endl;
      }else{
        for(int i = 1; i <= N; i++){
          cout << i << " ";
        }
        cout << endl;
      }
    }else{
      cout << -1 << endl;
    }
  }
  else{

    vector<int> upper;
    vector<int> lower;
    for(int i = N - A + 1; i <= N; i++){
      upper.push_back(i);
    }
    for(int i = B-1; i <= N-A; i++){
        lower.push_back(i);
      }
      for(int i = B-2; i >= 1; i--){
        lower.push_back(i);
    }
    
    vector<int> ans;

    int j, k;
    j = 0; k = 0;
    while(j < upper.size() || k < lower.size()){
      if(j < upper.size()){
        ans.push_back(upper[j]);
        j++;
      }
      if(k < lower.size()){
        ans.push_back(lower[k]);
        k++;
      }
    }

    if(is_swaped){
      for(int i = N-1; i >= 0; i--){
        cout << ans[i] << " ";
      }
      cout << endl;
    }else{
      for(int i = 0; i < N; i++){
        cout << ans[i] << " ";
      }
      cout << endl;
    }


  }
  


  return 0;
}