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>

using namespace std;

#define int long long
// #define double long double
#define FOR(i, a, b) for(ll i = (a); i < (b); ++i)
#define FORR(i, a, b) for(ll i = (a); i > (b); --i)
#define REP(i, n) for(ll i = 0; i < (n); ++i)
#define REPR(i, n) for(ll i = n; i >= 0; i--)
#define FOREACH(x, a) for(auto &(x) : (a))
#define VECCIN(x)                                                              \
    for(auto &youso_ : (x)) cin >> youso_
#define bitcnt(x) __builtin_popcount(x)
#define lbit(x) __builtin_ffsll(x)
#define rbit(x) __builtin_clzll(x)
#define SZ(x) ((ll)(x).size())
#define fi first
#define se second
#define All(a) (a).begin(), (a).end()
#define rAll(a) (a).rbegin(), (a).rend()

template <typename T = long long> inline T IN() {
    T x;
    cin >> x;
    return (x);
}
inline void CIN() {}
template <class Head, class... Tail>
inline void CIN(Head &&head, Tail &&... tail) {
    cin >> head;
    CIN(move(tail)...);
}
#define CCIN(...)                                                              \
    char __VA_ARGS__;                                                          \
    CIN(__VA_ARGS__)
#define DCIN(...)                                                              \
    double __VA_ARGS__;                                                        \
    CIN(__VA_ARGS__)
#define LCIN(...)                                                              \
    ll __VA_ARGS__;                                                            \
    CIN(__VA_ARGS__)
#define SCIN(...)                                                              \
    string __VA_ARGS__;                                                        \
    CIN(__VA_ARGS__)
#define Yes(a) cout << (a ? "Yes" : "No") << "\n"
#define YES(a) cout << (a ? "YES" : "NO") << "\n"
#define Printv(v)                                                              \
    {                                                                          \
        FOREACH(x, v) { cout << x << " "; }                                    \
        cout << "\n";                                                          \
    }
template <typename T = string> inline void eputs(T s) {
    cout << s << "\n";
    exit(0);
}
template <typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val) {
    std::fill((T *)array, (T *)(array + N), val);
}

template <typename T> using PQG = priority_queue<T, vector<T>, greater<T>>;
template <typename T> using PQ = priority_queue<T>;

typedef long long ll;
typedef vector<ll> VL;
typedef vector<VL> VVL;
typedef pair<ll, ll> PL;
typedef vector<PL> VPL;
typedef vector<bool> VB;
typedef vector<double> VD;
typedef vector<string> VS;

const int INF = 1e9;
const int MOD = 1e9 + 7;
// const int MOD = 998244353;
const ll LINF = 1e18;
// const double PI = atan(1.0) * 4.0;
const ll dx[] = {1, 1, 0, -1, -1, -1, 0, 1};
const ll dy[] = {0, 1, 1, 1, 0, -1, -1, -1};
#define PI 3.141592653589793238

signed main() {
    LCIN(N, A, B);
    if(A + B > N + 1 || A * B < N) eputs(-1);
    VVL ans(A);
    REP(i, B) { ans[0].push_back(i + 1); }
    FOR(i, B, N) { ans[i % (A - 1) + 1].push_back(i + 1); }
    REP(i, A) REPR(j, ans[i].size() - 1) { cout << ans[i][j] << " "; }
    cout << "\n";
}