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: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<ll> vi;
typedef pair<int,ll> pi;
typedef vector<pi> vpi;
typedef double dl;
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define MOD 1e9+7
#define lb lower_bound
#define ub upper_bound
#define dst distance
#define INF 1000000000
#define ALL(x) x.begin(), x.end()
#define SZ(x) (int)x.size()
#define MAXN 300100

int N,U,D;

vi V;
bool done[MAXN];

int main(){
	cin>>N>>U>>D;
	if (U*D<N||U+D>N+1){
		cout<<-1;
		return 0;
	}
	for (int i=N;i>0;i-=D){V.pb(i);done[i]=1;}
	for (int i=1;i<=N&&SZ(V)<U;++i){
		if(done[i])continue;
		V.pb(i);
		done[i]=1;
	}
	vi out;
	sort(ALL(V));
	for (auto i : V){
		int t = i-1;
		out.pb(i);
		while(t>0&&!done[t]){
			out.pb(t);
			--t;
			done[t]=1;
		}
	}
	assert(SZ(out) == N);
	for (auto i : out)cout<<i<<' ';
}