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<iostream>
#include<iomanip>
#include<math.h>
#include<vector>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<bitset>
#include<random>
#include<time.h>
#define INF 1000000000ll
#define MOD 1000000007ll
#define EPS 1e-10
#define REP(i,m) for(long long i=0; i<m; i++)
#define FOR(i,n,m) for(long long i=n; i<m; i++)
#define DUMP(a) for(long long dump=0; dump<(ll)a.size(); dump++) { cout<<a[dump]; if(dump!=(ll)a.size()-1) cout<<" "; else cout<<endl; }
#define ALL(v) v.begin(),v.end()
#define pb push_back
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
typedef long double ld;

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	ll n,a,b;
	cin>>n>>a>>b;
	vector<ll> ans(n);
	if(a*b<n) {
		cout<<-1<<endl;
		return 0;
	}
	vector<ll> cut(b);
	ll nokori=n;
	REP(i,b) {
		if(nokori>b-1-i) {
			cut[i]=min(a,nokori-(b-1-i));
			nokori-=cut[i];
		}
	}
	ll sum=0;
	ll idx=0;
	REP(i,b) {
		REP(j,cut[i]) {
			ans[idx]=n-sum-(cut[i]-1-j);
			idx++;
		}
		sum+=cut[i];
	}
	DUMP(ans);
}