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 mod = 1e9 + 7;
int cur;
struct Matrix {
  long long a[310][310];
} a;
Matrix mul(Matrix a, Matrix b) {
  Matrix res;
  memset(res.a, 0, sizeof res.a);
  for (int i = 0; i < cur; i++)
    for (int j = 0; j < cur; j++)
      for (int k = 0; k < cur; k++)
        (res.a[i][j] += a.a[i][k] * b.a[k][j] % mod) %= mod;
  return res;
}
Matrix pow(Matrix a, long long n) {
  Matrix ans, base = a;
  for (int i = 0; i < cur; i++) ans.a[i][i] = 1;
  while (n) {
    if (n & 1) ans = mul(ans, base);
    base = mul(base, base);
    n >>= 1;
  }
  return ans;
}
int main() {
  int n, k, m;
  scanf("%d%d%d", &n, &k, &m);
  cur = (k + 1) * (1 << m);
  for (int i = 0; i <= k; i++) {
    for (int j = 0; j < (1 << m); j++) {
      int nex = (j << 1) & ((1 << m) - 1), cnt = __builtin_popcount(j) + 1;
      a.a[i * (1 << m) + j][i * (1 << m) + nex] = 1;
      if (i < k) a.a[i * (1 << m) + j][(i + 1) * (1 << m) + (nex | 1)] = cnt;
    }
  }
  a = pow(a, n);
  long long ans = 0;
  for (int i = 0; i < (1 << m); i++) (ans += a.a[0][k * (1 << m) + i]) %= mod;
  printf("%lld\n", ans);
  return 0;
}
