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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>


    int writelist[300000];
    int unko;
    
int write(int N,int x){
    if(unko>=N){
        for(int i=0;i<N;i++){
            if(writelist[i]>N||writelist[i]<=0){printf("-1");return 1;}
        }
        for(int i=0;i<N;i++)printf("%d ",writelist[i]);
        return 1;
    }
    else{
        writelist[unko]=x;
        unko++;
        return 0;
    }
}

int main(void){
    unko=0;
    int N,A,B;
    scanf("%d %d %d",&N,&A,&B);
    if(A+B>N+1)printf("-1");
    else{
        int gensyo=0;
        for(int i=1;;i++){
            for(int j=N-A*i+1;j<=N-A*(i-1);j++){if(write(N,j)==1)return 0;}//printf("%d ",j);//TODO
            gensyo++;
            if(N-i*A>=B-gensyo&&N-(i+1)*A<B-(gensyo+1)){
                int leng=N-i*A-(B-gensyo-1);
                for(int j=N-i*A-leng+1;j<=N-i*A;j++){if(write(N,j)==1)return 0;}
                for(int j=N-i*A-leng;j>=1;j--){if(write(N,j)==1)return 0;}
                break;
            }
        }
    }
    write(N,0);
    return 0;
}