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: Accepted)

        #include <bits/stdc++.h>
        using namespace std;
		typedef long long int ll;

        int main() {
          ll N,A,B;
          cin>> N >> A >> B; 
          if( A+B -1 > N || A*B < N || A > N || B>N || A<1 || B<1 ){
            cout<< -1 <<endl;
          }
          else if(B==1){
            for(ll i=1; i<=N; i++){
              cout<< i <<" "; 
            }
          }
          else{
            ll T = (N - A )/(B-1);
            ll R = (N - A)%(B-1);
            ll X;
            for(ll i=1; i<= T; i++){
              for(ll j=1; j<= B; j++){
                	cout<< B*i+ + 1 -j <<" ";  
              }
              X= B*i; 
            }
            if(R==0){
            	for(ll i=X+1; i<=N; i++){
              	cout<< i <<" ";
                }
            }else{
              for(ll i=X+1; i<= N-R-1; i++){
                cout<< i <<" ";
              }
              for(ll i=N; i>= N-R; i--){
                cout<< i <<" ";
              }
            }
          }
        }