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>

#define fr first
#define sc second
#define mk make_pair

using namespace std;

int n, a, b, u[1000001];

int main()
{
    cin >> n >> a >> b;

    if(a == 1)
    {
        if(b == n)
        {
            for(int i = n; i >= 1; i--)
                cout << i << " ";
        }else
        {
            cout << -1;
        }
        return 0;
    }
    if(b == 1)
    {
        if(a == n)
        {
            for(int i = 1; i <= n; i++)
                cout << i << " ";
        }else
        {
            cout << -1;
        }
        return 0;
    }

    if(a + b - 1 > n || a == 0 || b == 0)
    {
        cout << -1;
        return 0;
    }
    vector<int> vec;
    int l = b, r = n - a;
    for(int i = r + 1; i <= n; i++)
        vec.push_back(i);
    for(int i = l - 1; i >= 1; i--)
        vec.push_back(i);

    vector<int> ans = vec;

    a--;
    b--;
    int q = a + 1, w = b, e = 1, t = 1;
    while(ans.size() < n)
    {
        int fl1 = 0, fl2 = 0;
        for(int j = 1; j < vec.size() && ans.size() < n; j++)
        {
            if(j <= a)
            {
                if(fl1 == 0)
                    fl1 = 1, e++;
                ans.push_back(vec[j] - q);
            }else
            {
                if(fl2 == 0)
                    fl2 = 1, t++;
                ans.push_back(vec[j] + w);
            }
        }
        q += a;
        w += b;
    }
    if(e <= t)
        e++;
    else
        t++;
    if(e > a + 1 || t > b + 1)
    {
        cout << -1;
        return 0;
    }

    l = n, r = 1;
    for(int i = 0; i < ans.size(); i++)
    {
        u[ans[i]] = 1;
    }

    for(int i = 1; i <= n; i++)
    {
        if(u[i] == 0)
        {
            l = min(l, i);
            r = max(r, i);
        }
    }

    if(l <= r)
    {
        int y = r - l + 1;
        for(int i = ans.size() - 1; i >= ans.size() - y; i--)
        {
            ans[i] += y;
        }
    }
    for(int i = 0; i < ans.size(); i++)
        cout << ans[i] << " ";

    return 0;
}