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;
inline int gi() {
  int data = 0, fu = 1;
  char ch = 0;
  while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-') fu = -1, ch = getchar();
  while ('0' <= ch && ch <= '9')
    data = (data << 1) + (data << 3) + (ch ^ 48), ch = getchar();
  return data * fu;
}
int f[15][15][105], n, m, nL, K, Ans, len[15], inv[15], coef[15][105];
inline void add(int &x, int y) {
  x += y;
  if (x >= 1000000007) x -= 1000000007;
}
void Dfs(int nw, int res, int l) {
  add(coef[nw][len[nw]], res);
  if (nw == K) return;
  for (int d = 1; d <= m; ++d) {
    len[nw + 1] = len[nw] + d;
    while (len[l] + m < len[nw + 1]) ++l;
    Dfs(nw + 1, 1ll * res * (nw - l + 2) % 1000000007, l);
  }
  return;
}
inline int Comb(int N, int M) {
  if (M > N || N < 0 || M < 0) return 0;
  int ret = 1;
  for (int l = N - M + 1; l <= N; ++l) ret = 1ll * l * ret % 1000000007;
  for (int l = 1; l <= M; ++l) ret = 1ll * ret * inv[l] % 1000000007;
  return ret;
}
int main() {
  n = gi();
  K = gi();
  m = gi();
  len[1] = 1;
  Dfs(1, 1, 1);
  f[0][0][0] = 1;
  nL = m * (K - 1) + m * K;
  for (int i = 0; i <= K; ++i)
    for (int j = 0; j <= K; ++j)
      for (int l = 0; l <= nL; ++l)
        if (f[i][j][l])
          for (int s = K - j; s >= 1; --s)
            for (int d = nL - l - m * (i > 0); d >= 1; --d)
              add(f[i + 1][j + s][l + d + m * (i > 0)],
                  1ll * coef[s][d] * f[i][j][l] % 1000000007);
  Ans = 0;
  inv[0] = inv[1] = 1;
  for (int i = 2; i <= K; ++i)
    inv[i] =
        1ll * (1000000007 - 1000000007 / i) * inv[1000000007 % i] % 1000000007;
  for (int i = 1; i <= K; ++i)
    for (int l = 0; l <= nL; ++l)
      if (f[i][K][l])
        add(Ans, 1ll * f[i][K][l] * Comb(n - l + i, i) % 1000000007);
  cout << Ans << endl;
  return 0;
}
