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;
void add(int &x, int y) {
  x += y;
  if (x >= 1000000007) x -= 1000000007;
}
inline int read() {
  int x = 0;
  bool t = false;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') t = true, ch = getchar();
  while (ch <= '9' && ch >= '0') x = x * 10 + ch - 48, ch = getchar();
  return t ? -x : x;
}
int b[13][1 << 4], bel[1 << 4], n, K, m, N, ans;
struct Matrix {
  int s[210][210];
  void clear() {
    for (int i = 1; i <= N; ++i)
      for (int j = 1; j <= N; ++j) s[i][j] = 0;
  }
  void init() {
    clear();
    for (int i = 1; i <= N; ++i) s[i][i] = 1;
  }
  int *operator[](int x) { return s[x]; }
} A, Tr;
Matrix operator*(Matrix &a, Matrix &b) {
  Matrix c;
  c.clear();
  for (int i = 1; i <= N; ++i)
    for (int j = 1; j <= N; ++j)
      for (int k = 1; k <= N; ++k)
        c[i][j] = (c[i][j] + 1ll * a[i][k] * b[k][j]) % 1000000007;
  return c;
}
int main() {
  n = read();
  K = read();
  m = read();
  for (int i = 1; i < 1 << m; ++i) bel[i] = bel[i >> 1] + (i & 1);
  for (int j = 0; j <= K; ++j)
    for (int S = 0; S < 1 << m; ++S) b[j][S] = ++N;
  for (int j = 0; j <= K; ++j)
    for (int S = 0; S < 1 << m; ++S) {
      int T = (S << 1) & ((1 << m) - 1);
      add(Tr[b[j][S]][b[j][T]], 1);
      if (j < K) add(Tr[b[j][S]][b[j + 1][T | 1]], bel[S] + 1);
    }
  A[1][1] = 1;
  while (n) {
    if (n & 1) A = A * Tr;
    Tr = Tr * Tr;
    n >>= 1;
  }
  for (int i = 0; i < 1 << m; ++i) add(ans, A[1][b[K][i]]);
  printf("%d\n", ans);
  return 0;
}
