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>
using namespace std;

#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(a) begin(a), end(a)
#define SZ(a) ((int)(a).size())
#define rep(i, n) for (int i = 0; i < n; i++)

#ifdef __DEBUG
#define debug if (true)
#else
#define debug if (false)
#endif

typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;

template <typename T> struct vector2 : public vector<vector<T>> { vector2(int n, int m, const T& k = T()) : vector<vector<T>>(n, vector<T>(m, k)) {} };
template <typename T> struct vector3 : public vector<vector2<T>> { vector3(int n, int m, int k, const T &l = T()) : vector<vector2<T>>(n, vector2<T>(m, k, l)) {} };

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, inc, dec;
  cin >> n >> inc >> dec;
  bool rev = inc < dec;
  if (rev) {
    debug puts("REV");
    swap(inc, dec);
  }
  if (dec == 0) {
    puts("-1");
    return 0;
  }
  if (inc + dec - 1 > n) {
    puts("-1");
    return 0;
  }
  if (1LL * inc * (inc + dec - 1) < n) {
    puts("-1");
    return 0;
  }
  int rem = n - inc;
  int last = 0;
  vi ans;
  for (int iter = 0; iter < inc; iter++) {
    int take = min(dec - 1, rem);
    rem -= take;
    for (int i = last + take + 1; i > last; i--) {
      debug printf("PB %d\n", i);
      ans.push_back(i);
    }
    last = last + take + 1;
  }
  if (rev) {
    for (int &i : ans) i = n + 1 - i;
  }
  assert(SZ(ans) == n);
  for (int i = 0; i < SZ(ans); i++) {
    printf("%d%c", ans[i], i == SZ(ans) - 1 ? '\n' : ' ');
  }
  return 0;
}