This problem is same as the previous one, but has larger constraints.

Aki is playing a new video game. In the video game, he will control Neko, the giant cat, to fly between planets in the Catniverse.

There are n planets in the Catniverse, numbered from 1 to n. At the beginning of the game, Aki chooses the planet where Neko is initially located. Then Aki performs k - 1 moves, where in each move Neko is moved from the current planet x to some other planet y such that:

  * Planet y is not visited yet. 
  * 1 ≤ y ≤ x + m (where m is a fixed constant given in the input) 



This way, Neko will visit exactly k different planets. Two ways of visiting planets are called different if there is some index i such that, the i-th planet visited in the first way is different from the i-th planet visited in the second way.

What is the total number of ways to visit k planets this way? Since the answer can be quite large, print it modulo 10^9 + 7.

Input

The only line contains three integers n, k and m (1 ≤ n ≤ 10^9, 1 ≤ k ≤ min(n, 12), 1 ≤ m ≤ 4) — the number of planets in the Catniverse, the number of planets Neko needs to visit and the said constant m.

Output

Print exactly one integer — the number of different ways Neko can visit exactly k planets. Since the answer can be quite large, print it modulo 10^9 + 7.

Examples

Input


3 3 1


Output


4


Input


4 2 1


Output


9


Input


5 5 4


Output


120


Input


100 1 2


Output


100

Note

In the first example, there are 4 ways Neko can visit all the planets:

  * 1 → 2 → 3 
  * 2 → 3 → 1 
  * 3 → 1 → 2 
  * 3 → 2 → 1 



In the second example, there are 9 ways Neko can visit exactly 2 planets:

  * 1 → 2 
  * 2 → 1 
  * 2 → 3 
  * 3 → 1 
  * 3 → 2 
  * 3 → 4 
  * 4 → 1 
  * 4 → 2 
  * 4 → 3 



In the third example, with m = 4, Neko can visit all the planets in any order, so there are 5! = 120 ways Neko can visit all the planets.

In the fourth example, Neko only visit exactly 1 planet (which is also the planet he initially located), and there are 100 ways to choose the starting planet for Neko.
#include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-9;
template <std::uint_fast64_t Modulus>
class modint {
  using u64 = std::uint_fast64_t;
  u64 a;

 public:
  constexpr modint(const u64 x = 0) noexcept : a(x % Modulus) {}
  constexpr 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;
  }
  modint &operator+=(const modint rhs) noexcept {
    a += rhs.a;
    if (a >= Modulus) {
      a -= Modulus;
    }
    return *this;
  }
  modint &operator-=(const modint rhs) noexcept {
    if (a < rhs.a) {
      a += Modulus;
    }
    a -= rhs.a;
    return *this;
  }
  modint &operator*=(const modint rhs) noexcept {
    a = a * rhs.a % Modulus;
    return *this;
  }
  modint &operator/=(modint rhs) noexcept {
    u64 exp = Modulus - 2;
    while (exp) {
      if (exp % 2) {
        *this *= rhs;
      }
      rhs *= rhs;
      exp /= 2;
    }
    return *this;
  }
};
using mint = modint<1000000007>;
long long m;
vector<mint> matmul(vector<mint> &dp, vector<vector<mint>> &mt) {
  vector<mint> ret(m, 0);
  for (long long i = (0); i < (m); i++)
    for (long long j = (0); j < (m); j++) ret[i] += mt[i][j] * dp[j];
  return ret;
}
vector<vector<mint>> update(vector<vector<mint>> &mt) {
  vector<vector<mint>> ret(m, vector<mint>(m, 0));
  for (long long i = (0); i < (m); i++)
    for (long long j = (0); j < (m); j++)
      for (long long k = (0); k < (m); k++) ret[i][j] += mt[i][k] * mt[k][j];
  return ret;
}
void matpow(vector<mint> &dp, vector<vector<mint>> &mt, long long k) {
  while (k) {
    if (k % 2) dp = matmul(dp, mt);
    mt = update(mt);
    k /= 2;
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, k, l;
  cin >> n >> k >> l;
  long long M = 1 << l;
  m = M * (k + 2);
  vector<vector<mint>> mt(M * (k + 2), vector<mint>(M * (k + 2), 0));
  for (long long i = (0); i < (k + 1); i++) {
    for (long long j = (0); j < (M); j++) {
      long long mask = (j << 1) & (M - 1);
      mt[i * M + mask][i * M + j] += 1;
      mt[(i + 1) * M + (mask | 1)][i * M + j] += __builtin_popcount(j) + 1;
    }
  }
  vector<mint> dp(M * (k + 2), 0);
  dp[0] = 1;
  matpow(dp, mt, n);
  mint ans = 0;
  for (long long i = (0); i < (M); i++) ans += dp[M * k + i];
  cout << ans.value() << endl;
  return 0;
}
