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 N = 13 * 16, M = 1e9 + 7;
int n, m, k;
struct zz {
  int a[N][N];
} A;
inline zz cf(zz x, zz y) {
  zz z;
  memset(z.a, 0, sizeof z.a);
  for (int i = 0; i < (1 << m) * (k + 1); i++)
    for (int j = 0; j < (1 << m) * (k + 1); j++)
      for (int l = 0; l < (1 << m) * (k + 1); l++)
        z.a[i][j] = (z.a[i][j] + (long long)x.a[i][l] * y.a[l][j]) % M;
  return z;
}
inline zz ksm(zz x, int y) {
  zz ans;
  memset(ans.a, 0, sizeof ans.a);
  for (int i = 0; i < (1 << m) * (k + 1); i++) ans.a[i][i] = 1;
  for (; y; y >>= 1, x = cf(x, x))
    if (y & 1) ans = cf(ans, x);
  return ans;
}
int main() {
  scanf("%d%d%d", &n, &k, &m);
  for (int i = 0; i < 1 << m; i++)
    for (int j = 0; j <= k; j++) {
      int num = 1;
      for (int l = 0; l < m; l++)
        if ((1 << l) & i) num++;
      A.a[(i >> 1) * (k + 1) + j][i * (k + 1) + j] = 1;
      if (j < k)
        A.a[(i >> 1 | (1 << (m - 1))) * (k + 1) + j + 1][i * (k + 1) + j] = num;
    }
  A = ksm(A, n);
  int ans = 0;
  for (int i = 0; i < 1 << m; i++) (ans += A.a[i * (k + 1) + k][0]) %= M;
  printf("%d\n", ans);
}
