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 <bits/stdc++.h>
//#include <math.h>
using namespace std;

#define INF 1.1e9
#define LINF 1.1e18
#define FOR(i,a,b) for (int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(),(v).end()
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define BIT(x,n) bitset<n>(x)
#define PI 3.14159265358979323846

typedef long long ll;
typedef pair<int,ll> P;
typedef pair<ll,P> PP;

//-----------------------------------------------------------------------------

int n,a,b;

bool check(vector<int> &v) {
	int i=0,cnt=1;
	int A=0,B=0;
	while(i<n) {
		while(i+1<n&&v[i+1]>v[i]) cnt++,i++;
		if(cnt>a) return false;
		A=max(A,cnt);
		cnt=1;
		while(i+1<n&&v[i+1]<v[i]) cnt++,i++;
		if(cnt>b) return false;
		B=max(B,cnt);
		cnt=1;
		i++;
	}
	if(A<a||B<b) return false;
	return true;
}

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

	cin>>n>>a>>b;
	/*if(a+b>n+1||(a==1&&b!=n)||(b==1&&a!=n)) {
		cout<<-1<<endl;
		return 0;
	}*/
	int ma=n,mi=1;
	int turn=+1,A=a-1,B=b-1;
	vector<int> v;
	REP(i,n) {
		if(turn==1) {
			v.pb(mi);
			mi++,A--;
			if(A==0) turn=-1,A=a-1;
		}
		else {
			v.pb(ma);
			ma--,B--;
			if(B==0) turn=1,B=b-1;
		}
	}
	if(!check(v)) cout<<-1<<endl;
	else {
		for(auto x:v) cout<<x<<' ';
		cout<<endl;
	}

	return 0;
}