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 mo = 1e9 + 7;
struct node {
  int mat[210][210];
} qwq;
int n, k, m;
inline int ha(int x, int y) { return x * (1 << m) + y; }
node ksm(int l) {
  if (l == 1) return qwq;
  node op = ksm(l / 2), op2;
  memset(op2.mat, 0, sizeof(op2.mat));
  for (int i = 0; i < (k + 1) * (1 << m); i++)
    for (int j = 0; j < (k + 1) * (1 << m); j++)
      for (int h = 0; h < (k + 1) * (1 << m); h++)
        op2.mat[i][j] =
            (op2.mat[i][j] + 1ll * op.mat[i][h] * op.mat[h][j]) % mo;
  if (l % 2 == 0) return op2;
  node op3;
  memset(op3.mat, 0, sizeof(op3.mat));
  for (int i = 0; i < (k + 1) * (1 << m); i++)
    for (int j = 0; j < (k + 1) * (1 << m); j++)
      for (int h = 0; h < (k + 1) * (1 << m); h++)
        op3.mat[i][j] =
            (op3.mat[i][j] + 1ll * qwq.mat[i][h] * op2.mat[h][j]) % mo;
  return op3;
}
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 ovo = (j << 1) % (1 << m);
      qwq.mat[ha(i, ovo)][ha(i, j)]++;
      if (i > 0) {
        int way_to_here = 1;
        for (int r = 0; r < m; r++)
          if ((1 << r) & j) way_to_here++;
        qwq.mat[ha(i, ovo | 1)][ha(i - 1, j)] += way_to_here;
      }
    }
  node uoj = ksm(n);
  int ans = 0;
  for (int i = 0; i < (1 << m); i++)
    ans = (ans + 1ll * uoj.mat[ha(k, i)][0]) % mo;
  printf("%d", ans);
  return 0;
}
