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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define REP(i,n) for (int i=0;i<(n);i++)
#define FOR(i,s,e) for (int i=s;i<(e);i++)
#define All(v) (v).begin(),(v).end()
#define mp(a,b) make_pair(a,b)
#define pb(a) push_back(a)
using namespace std;
typedef long long llint;
typedef pair<int, int> P;
const int MOD = (int)1e9 + 7;
const int INF = 999999999;

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

    llint N, A, B;
    cin >> N >> A >> B;
    if(N+1 < A+B) puts("-1");
    else if(A*B < N) puts("-1");
    else{
        vector<int> P;
        int now=N, ct=N-(A+B-1);
        REP(b, B){
            for(int a=A-1; a>=0; a--){
                if(a == A-1 || b == 0) P.pb(now-a);
                else if(ct > 0){
                    P.pb(now-a);
                    ct--;
                }
            }
            now -= A;
        }
        vector<int> Pst = P;
        sort(All(Pst));

        REP(i, N){
            cout << (lower_bound(All(Pst), P[i])-Pst.begin())+1 << (i == N-1 ? "\n" : " ");
        }
    }
    
    return 0;
}