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;
template <typename T>
bool chkmx(T &x, T y) {
  return x < y ? x = y, true : false;
}
template <typename T>
bool chkmn(T &x, T y) {
  return x > y ? x = y, true : false;
}
const int maxw = 250;
const int mod = 1e9 + 7;
int n, m, k, mask, lim, ans;
inline void upd(int &x) { x = (x >= mod) ? (x - mod) : x; }
struct Mat {
  int a[maxw][maxw];
  Mat() { memset(a, 0, sizeof a); }
  inline int *operator[](int x) { return a[x]; }
  inline friend Mat operator*(Mat x, Mat y) {
    Mat ans;
    for (int i = (0); i <= int(lim); i++)
      for (int j = (0); j <= int(lim); j++)
        for (int k = (0); k <= int(lim); k++)
          upd(ans[i][j] += 1ll * x[i][k] * y[k][j] % mod);
    return ans;
  }
} x, y;
Mat fpw(Mat a, int b) {
  Mat ans = a;
  b--;
  for (; b; b >>= 1, a = a * a)
    if (b & 1) ans = ans * a;
  return ans;
}
inline int id(int i, int j) { return i * (mask + 1) + j; }
int main() {
  scanf("%d%d%d", &n, &k, &m), mask = (1 << m) - 1,
                               lim = (k + 1) * (mask + 1) - 1;
  for (int i = (0); i <= int(mask); i++)
    for (int j = (0); j <= int(k - 1); j++)
      y[id(j, i)][id(j + 1, (1 | ((i << 1) & mask)))] =
          1 + __builtin_popcount(i);
  for (int i = (0); i <= int(mask); i++)
    for (int j = (0); j <= int(k); j++)
      y[id(j, i)][id(j, ((i << 1) & mask))] = 1;
  x[0][0] = 1, x = x * fpw(y, n);
  for (int i = (id(k, 0)); i <= int(lim); i++) upd(ans += x[0][i]);
  printf("%d\n", ans);
  return 0;
}
