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>
const int P = 1e9 + 7;
int n, k, m;
void add(int &x, int y) { (x += y) >= P && (x -= P); }
struct Matrix {
  int n, m;
  std::vector<std::vector<int>> A;
  Matrix(int n, int m) : n(n), m(m) { A.assign(n, std::vector<int>(m, 0)); }
  void identity() {
    for (int i = 0; i < n; i++) A[i][i] = 1;
  }
  friend Matrix operator*(Matrix a, Matrix b) {
    Matrix c(a.n, b.m);
    for (int i = 0; i < a.n; i++) {
      for (int j = 0; j < b.m; j++) {
        for (int k = 0; k < a.m; ++k) {
          add(c.A[i][j], 1LL * a.A[i][k] * b.A[k][j] % P);
        }
      }
    }
    return c;
  }
  friend Matrix operator^(Matrix x, int p) {
    Matrix ans(x.n, x.n);
    ans.identity();
    for (; p; p >>= 1, x = x * x) {
      if (p & 1) ans = ans * x;
    }
    return ans;
  }
};
int id(int i, int s) { return i * (1 << m) + s; }
int main() {
  scanf("%d%d%d", &n, &k, &m);
  int msk = 1 << m, tot = (k + 1) * msk;
  Matrix x(tot, tot), f(tot, 1);
  for (int i = 0; i <= k; i++) {
    for (int cur = 0; cur < msk; cur++) {
      int nxt = (cur << 1) % msk;
      x.A[id(i, nxt)][id(i, cur)] = 1;
      if (i < k) {
        int v = __builtin_popcount(cur) + 1;
        x.A[id(i + 1, nxt | 1)][id(i, cur)] = v;
      }
    }
  }
  f.A[0][0] = 1;
  f = (x ^ n) * f;
  int ans = 0;
  for (int i = 0; i < msk; i++) add(ans, f.A[id(k, i)][0]);
  printf("%d\n", ans);
  return 0;
}
