output_description: 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.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N A B
attempt: //header
  #define set_header 1
  #ifdef set_header
      #include <bits/stdc++.h>
      #ifdef LOCAL
          #include "cxx-prettyprint-master/prettyprint.hpp"
          #define print(x) cout << x << endl
      #else
          #define print(...) 42
      #endif
      using namespace std;
      using ll = long long;
      #define INF 1'010'000'000'000'000'017LL
      #define mod 998244353LL
      #define eps 0.0001
      #define all(x) (x).begin(), (x).end()
      #define reverse(x) (x).rbegin(), (x).rend()
      #define FOR(i,a,b) for(ll i=(a);i<(b);++i)
      #define rep(i,n)  FOR(i,0,n)
      #define SZ(x) ((ll)(x).size())
      #define sum(x) accumulate(ALL(x), 0LL)//?
      #define pb(x) push_back(x)
      typedef pair < ll , ll >P;
      ll gcd(ll a, ll b) { return b ? gcd(b, a%b) : a; }
      template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
      template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
      template <std::uint_fast64_t Modulus> class modint {
          using u64 = std::uint_fast64_t;
  
          public:
          u64 a;
  
          constexpr modint(const u64 x = 0) noexcept : a(x % Modulus) {}
          constexpr u64 &value() noexcept { return a; }
          constexpr const u64 &value() const noexcept { return a; }
          constexpr modint operator+(const modint rhs) const noexcept {
              return modint(*this) += rhs;
          }
          constexpr modint operator-(const modint rhs) const noexcept {
              return modint(*this) -= rhs;
          }
          constexpr modint operator*(const modint rhs) const noexcept {
              return modint(*this) *= rhs;
          }
          constexpr modint operator/(const modint rhs) const noexcept {
              return modint(*this) /= rhs;
          }
          constexpr modint &operator+=(const modint rhs) noexcept {
              a += rhs.a;
              if (a >= Modulus) {
              a -= Modulus;
              }
              return *this;
          }
          constexpr modint &operator-=(const modint rhs) noexcept {
              if (a < rhs.a) {
              a += Modulus;
              }
              a -= rhs.a;
              return *this;
          }
          constexpr modint &operator*=(const modint rhs) noexcept {
              a = a * rhs.a % Modulus;
              return *this;
          }
          constexpr modint &operator/=(modint rhs) noexcept {
              u64 exp = Modulus - 2;
              while (exp) {
              if (exp % 2) {
                  *this *= rhs;
              }
              rhs *= rhs;
              exp /= 2;
              }
              return *this;
          }
          };
      template< typename T >
      T mod_pow(T x, T n, const T &p) {
          T ret = 1;
          while(n > 0) {
              if(n & 1) (ret *= x) %= p;
              (x *= x) %= p;
              n >>= 1;
          }
          return ret;
      }
      class range {private: struct I{int x;int operator*(){return x;}bool operator!=(I& lhs){return x<lhs.x;}void operator++(){++x;}};I i,n;
      public:range(int n):i({0}),n({n}){}range(int i,int n):i({i}),n({n}){}I& begin(){return i;}I& end(){return n;}};
      uint64_t my_rand(void) {
  	static uint64_t x = 88172645463325252ULL;
  	x = x ^ (x << 13); x = x ^ (x >> 7);
  	return x = x ^ (x << 17);
  }
  #endif
  //library
  
  //main
  int main() {
      int N, A, B;
      cin >> N >> A >> B;
      if(N+1<A+B || N>A*B){
          cout << -1 << endl;
          return 0;
      }
      int tmp = N;
      vector<int> ans(0);
      for(int i = B;i>0;i--)ans.pb(i);
      tmp-=B;
      int now = B+1;
      for(int a = A-1;a>0;a--){
          int q = tmp/a;
          tmp -=q;
          for(int b = now+q-1;b>=now;b--) ans.pb(b);
          now+=q;
      }
      rep(i, N) cout << ans[i] <<' ';
      cout << endl;
  }
problem_description: 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.
samples: [{"input": "5 3 2", "output": "2 4 1 5 3\n    \n\nOne longest increasing subsequence of this sequence is {2,4,5}, and one\nlongest decreasing subsequence of it is {4,3}.\n\n* * *"}, {"input": "7 7 1", "output": "1 2 3 4 5 6 7\n    \n\n* * *"}, {"input": "300000 300000 300000", "output": "-1"}]