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 P = 1e9 + 7;
int K, n, i, j, nn, ans, m, k;
inline void add(int &x, int y) {
  x += y;
  if (x >= P) x -= P;
}
struct M {
  int a[208][208];
  friend M operator*(M x, M y) {
    M z;
    memset(z.a, 0, sizeof(z.a));
    for (int i = 0; i < nn; i++)
      for (int j = 0; j < nn; j++)
        for (int k = 0; k < nn; k++)
          add(z.a[i][j], 1ll * x.a[i][k] * y.a[k][j] % P);
    return z;
  }
  friend M operator^(M x, int y) {
    M z;
    memset(z.a, 0, sizeof(z.a));
    z.a[0][0] = 1;
    for (; y; y >>= 1, x = x * x)
      if (y & 1) z = z * x;
    return z;
  }
} A, dp;
int id(int x, int y) { return x * (1 << m) + y; }
int main() {
  scanf("%d%d%d", &n, &K, &m);
  for (i = 0; i <= K; i++)
    for (j = 0; j < 1 << m; j++) {
      k = (j << 1) & ((1 << m) - 1);
      A.a[id(i, j)][id(i, k)] = 1;
      if (i < K) A.a[id(i, j)][id(i + 1, k | 1)] = __builtin_popcount(j) + 1;
    }
  nn = id(K, 1 << m);
  A = A ^ n;
  for (i = 0; i < 1 << m; i++) add(ans, A.a[0][id(K, i)]);
  printf("%d", ans);
}
