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)

#pragma comment(linker, "/stack:200000000")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")


#include <bits/stdc++.h>
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// #include <ext/pb_ds/detail/standard_policies.hpp>

using namespace std;
// using namespace __gnu_pbds;
// using namespace __gnu_cxx;

typedef long long ll;
typedef unsigned int ul;
typedef unsigned long long ull;
typedef vector <int> vi;
typedef map<int, vector<int> > mvii;
typedef map<int, int> mii;
typedef queue <int> qi;
typedef vector <string> vs;
typedef pair <int, int> pii;
typedef vector<pair<int,int> > vpii;

// Order Statistic Tree

/* Special functions: 

		find_by_order(k) --> returns iterator to the kth largest element counting from 0
		order_of_key(val) --> returns the number of items in a set that are strictly smaller than our item
*/

// typedef tree<
// int,
// null_type,
// less<int>,
// rb_tree_tag,
// tree_order_statistics_node_update>
// ordered_set;

#define MP make_pair
#define SORT(a) sort (a.begin(), a.end())
#define REVERSE(a) reverse (a.begin(), a.end())
#define ALL(a) a.begin(), a.end()
#define PI acos(-1)
#define ms(x,y) memset (x, y, sizeof (x))
#define inf 1e9
#define INF 1e16
#define pb push_back
#define MAX 100005
#define debug(a,b) cout<<a<<": "<<b<<endl
#define Debug cout<<"Reached here"<<endl
#define prnt(a) cout<<a<<"\n"
#define mod 1000000007LL
#define FOR(i,a,b) for (int i=(a); i<(b); i++)
#define FORr(i,a,b) for (int i=(a); i>=(b); i--)
#define itrALL(c,itr) for(__typeof((c).begin()) itr=(c).begin();itr!=(c).end();itr++)
#define lc ((node)<<1)
#define rc ((node)<<1|1)
#define VecPrnt(v) FOR(J,0,v.size()) cout<<v[J]<<" "; cout<<endl
#define endl "\n"
#define PrintPair(x) cout<<x.first<<" "<<x.second<<endl
#define EPS 1e-9
#define ArrPrint(a,st,en) for(int J=st; J<=en; J++) cout<<a[J]<<" "; cout<<endl;

/* Direction Array */

// int fx[]={1,-1,0,0};
// int fy[]={0,0,1,-1};
// int fx[]={0,0,1,-1,-1,1,-1,1};
// int fy[]={-1,1,0,0,1,1,-1,-1};

template <class T> inline T bigmod(T p,T e,T M)
{
    ll ret = 1;
    for(; e > 0; e >>= 1)
    {
        if(e & 1) ret = (ret * p) % M;
        p = (p * p) % M;
    } return (T)ret;
}

template <class T> inline T gcd(T a,T b){if(b==0)return a;return gcd(b,a%b);}
template <class T> inline T modinverse(T a,T M){return bigmod(a,M-2,M);}
template <class T> inline T lcm(T a,T b) {a=abs(a);b=abs(b); return (a/gcd(a,b))*b;}
template <class T, class X> inline bool getbit(T a, X i) { T t=1; return ((a&(t<<i))>0);}
template <class T, class X> inline T setbit(T a, X i) { T t=1;return (a|(t<<i)); }
template <class T, class X> inline T resetbit(T a, X i) { T t=1;return (a&(~(t<<i)));}

inline ll getnum()
{
    char c = getchar();
    ll num,sign=1;
    for(;c<'0'||c>'9';c=getchar())if(c=='-')sign=-1;
    for(num=0;c>='0'&&c<='9';)
    {
        c-='0';
        num = num*10+c;
        c=getchar();
    }
    return num*sign;
}

inline ll power(ll a, ll b)
{
	ll multiply=1;
	FOR(i,0,b)
	{
		multiply*=a;
	}
	return multiply;
}

/****** END OF HEADER ******/

vi take[MAX*3];

int main()
{
    // ios_base::sync_with_stdio(0);
    // cin.tie(NULL); cout.tie(NULL);
    // freopen("in.txt","r",stdin);

    int test, cases=1;
    ll n, a, b;

    cin>>n>>a>>b;

    if(n>a*b || a+b-1>n) prnt(-1);
    else
    {
    	int rest=n, val=1, k;

    	FOR(i,0,a)
    	{
    		if(i==0) 
    		{
				while(take[i].size()<b) 
				{
					take[i].pb(val);
					val++;
				}
    			rest-=b;
    			k=rest/(a-1);

    			// cout<<rest<<" "<<a-1<<endl;
    			// VecPrnt(take[0]);
    			// prnt(k);
    		}
    		else
    		{
    			// prnt(min(rest,k));
    			
    			while(take[i].size()<min(k,rest))
    			{
    				take[i].pb(val);
    				val++;
    			}

    			// VecPrnt(take[i]);

    			rest-=take[i].size();
    		}
    	}

    	// prnt(rest);

    	FOR(i,1,rest+1) take[a-1].pb(val++);

    	FOR(i,0,a)
    	{
    		FORr(j,take[i].size()-1,0)
    			printf("%d ", take[i][j]);
    	}
		puts("");
    }

    return 0;
}