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 int mod = 1000000007;
inline int add(int a, int b) {
  if ((a += b) >= mod) a -= mod;
  return a;
}
inline int mult(int a, int b) {
  long long t = 1ll * a * b;
  if (t >= mod) t %= mod;
  return t;
}
namespace matrix {
template <int N, int M>
class matr {
 public:
  int dat[N][M];
  void init() { memset(dat, 0, sizeof(dat)); }
  void iden() {
    init();
    for (int i = 0; i < min(N, M); i++) dat[i][i] = 1;
  }
  matr<M, N> trans() {
    matr<M, N> out;
    for (int i = 0; i < M; i++)
      for (int j = 0; j < N; j++) out[i][j] = dat[j][i];
  }
  const int *operator[](const int &x) const { return dat[x]; }
  int *operator[](const int &x) { return dat[x]; }
  matr<N, M> operator+(const matr<N, M> a) {
    matr<N, M> out;
    for (int i = 0; i < N; i++)
      for (int j = 0; j < M; j++) out[i][j] = add(dat[i][j], a[i][j]);
    return out;
  }
  matr<N, M> operator*(const int &x) {
    matr<N, M> out = *this;
    for (int i = 0; i < N; i++)
      for (int j = 0; j < M; j++) out[i][j] = mult(out[i][j], x);
    return out;
  }
  template <int X>
  friend matr<N, M> operator*(const matr<N, X> a, const matr<X, M> b) {
    matr<N, M> out;
    out.init();
    for (int i = 0; i < N; i++) {
      for (int j = 0; j < X; j++) {
        if (a[i][j]) {
          for (int k = 0; k < M; k++)
            out[i][k] = add(out[i][k], mult(a[i][j], b[j][k]));
        }
      }
    }
    return out;
  }
};
};  // namespace matrix
using namespace matrix;
int n, K, m;
matr<13 * 16, 13 * 16> trans;
inline int getid(int i, int j) { return (i << m) | j; }
int main() {
  scanf("%d%d%d", &n, &K, &m);
  for (int i = 0; i <= K; i++) {
    for (int j = 0; j < (1 << m); j++) {
      int S1 = getid(i, j);
      int curs = (j >> 1);
      int S2 = getid(i, curs);
      trans[S1][S2] = add(trans[S1][S2], 1);
      if (i + 1 <= K) {
        curs = (j >> 1) | (1 << (m - 1));
        S2 = getid(i + 1, curs);
        trans[S1][S2] = add(trans[S1][S2], __builtin_popcount(j) + 1);
      }
    }
  }
  matr<1, 13 * 16> ans;
  ans.init();
  ans[0][getid(0, 0)] = 1;
  while (n) {
    if (n & 1) ans = ans * trans;
    trans = trans * trans;
    n >>= 1;
  }
  int ret = 0;
  for (int i = 0; i < (1 << m); i++) ret = add(ret, ans[0][getid(K, i)]);
  printf("%d\n", ret);
  return 0;
}
