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;
namespace thecold {
int n, k, m, lim = 192;
const int jzp = 1e9 + 7;
const int Max_n = 250;
int tot;
struct Matrix {
  int h[Max_n][Max_n];
  inline int* operator[](const int a) { return h[a]; }
  inline void clear() { memset(h, 0, sizeof(h)); }
  inline void init() {
    clear();
    for (int i = 1; i <= tot; ++i) h[i][i] = 1;
  }
} chg, ans;
inline int add(const int x, const int y) {
  int ret = x + y;
  if (ret >= jzp) ret -= jzp;
  return ret;
}
inline void ADD(int& x, const int y) {
  x += y;
  if (x >= jzp) x -= jzp;
}
inline Matrix operator*(Matrix a, Matrix b) {
  Matrix ret;
  ret.clear();
  for (int i = 1; i <= tot; ++i)
    for (int k = 1; k <= tot; ++k)
      for (int j = 1; j <= tot; ++j)
        ADD(ret[i][j], 1ll * a[i][k] * b[k][j] % jzp);
  return ret;
}
inline Matrix operator^(Matrix x, int y) {
  Matrix ret;
  ret.init();
  while (y) {
    if (y & 1) ret = ret * x;
    x = x * x, y >>= 1;
  }
  return ret;
}
const int Max_m = 15, Max_s = (1 << 4) + 5;
int rev[Max_m][Max_s], cnt[Max_s];
inline void main() {
  scanf("%d %d %d", &n, &k, &m);
  int all = (1 << m) - 1;
  for (int i = 1; i <= all; ++i) cnt[i] = cnt[i >> 1] + (i & 1);
  for (int i = 0; i <= k; ++i)
    for (int s = 0; s <= all; ++s) rev[i][s] = ++tot;
  for (int i = 0; i <= k; ++i)
    for (int s = 0; s <= all; ++s) {
      int now = rev[i][s], nex = rev[i][(s << 1) & all];
      chg[now][nex] = 1;
      nex = rev[i + 1][(s << 1 | 1) & all];
      if (nex) chg[now][nex] = 1 + cnt[s];
    }
  chg = chg ^ n;
  int ret = 0;
  for (int s = 0; s <= all; ++s) ADD(ret, chg[1][rev[k][s]]);
  printf("%d\n", ret);
}
}  // namespace thecold
int main() {
  thecold::main();
  return 0;
}
