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 = 1000000007;
inline void Add(int &a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
int sz;
struct Matrix {
  int mat[250][250];
  Matrix(int t = 0) {
    for (int i = 0; i < sz; i++)
      for (int j = 0; j < sz; j++) mat[i][j] = 0;
    for (int i = 0; i < sz; i++) mat[i][i] = t;
  }
};
Matrix operator*(const Matrix &a, const Matrix &b) {
  Matrix c;
  for (int i = 0; i < sz; i++)
    for (int j = 0; j < sz; j++)
      for (int k = 0; k < sz; k++)
        Add(c.mat[i][j], 1ll * a.mat[i][k] * b.mat[k][j] % mod);
  return c;
}
int n, k, m;
Matrix a, res;
int main() {
  scanf("%d%d%d", &n, &k, &m);
  sz = (k + 1) * (1 << m);
  res.mat[0][0] = 1;
  int C = (1 << m) - 1;
  for (int j = 0; j <= k; j++) {
    for (int mask = 0; mask < (1 << m); mask++) {
      a.mat[j * (1 << m) + ((mask << 1) & C)][j * (1 << m) + mask] = 1;
      if (j < k) {
        a.mat[(j + 1) * (1 << m) + (((mask << 1) & C) ^ 1)]
             [j * (1 << m) + mask] = __builtin_popcount(mask) + 1;
      }
    }
  }
  while (n) {
    if (n & 1) res = a * res;
    a = a * a;
    n >>= 1;
  }
  int ans = 0;
  for (int mask = 0; mask < (1 << m); mask++)
    Add(ans, res.mat[k * (1 << m) + mask][0]);
  printf("%d\n", ans);
  return 0;
}
