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<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<set>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<map>

#define Rep(i,x,y) for (int i=x,_Lim=y;i<=_Lim;i++)
#define Dep(i,x,y) for (int i=x,_Lim=y;i>=_Lim;i--)

#define mk(a,b) make_pair(a,b)

using namespace std;

typedef long long ll;
typedef pair<int,int> PII;

int Stv[1<<8];

inline int read(){
	int f=1,res=0,ch=getchar();
	while (ch<'0' || ch>'9'){
		if (ch=='-') f=-1;ch=getchar();
	}
	while (ch>='0' && ch<='9') res=res*10+ch-48,ch=getchar();
	return f*res;
}
int n,A,B;
int main(){
	n=read(),A=read(),B=read();
	if (A+B>n+1 || 1ll*A*B<n){
		puts("-1");return 0;
	}
	Rep(len,0,len){
		int kes=(n-len-1)/A+1;
		if (kes+len==B){
//			printf("kes=%d len=%d\n",kes,len);
			Rep(i,1,kes){
				int lim=(i==kes)?(n-len-(kes-1)*A):A;
				Rep(j,1,lim){
					printf("%d ",n-(i-1)*A-lim+j);
				}
			}
			Rep(i,1,len) printf("%d ",i);puts("");
			return 0;
		}
	}
}