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>
#pragma GCC optimize("Ofast")
using namespace std;
const int mod = 1000000007;
inline void inc(int &x, int y) { (x += y) >= mod && (x -= mod); }
inline int mul(int x, int y) { return 1LL * x * y - 1LL * x * y / mod * mod; }
int n, k, m, ans;
struct matrix {
  int mat[225][225];
  void clear() { memset(mat, 0, sizeof(mat)); }
  matrix(int x = 0) {
    clear();
    for (int i = 0; i < 225; ++i) mat[i][i] = x;
  }
  int *operator[](const int &x) { return mat[x]; }
  friend matrix operator*(matrix &a, matrix &b) {
    matrix res;
    for (int i = 0; i < 225; ++i)
      for (int j = 0; j < 225; ++j)
        for (int k = 0; k < 225; ++k) inc(res[i][j], mul(a[i][k], b[k][j]));
    return res;
  }
  friend matrix operator^(matrix a, long long b) {
    matrix res(1);
    while (b) {
      if (b & 1) res *= a;
      b >>= 1, a *= a;
    }
    return res;
  }
  matrix &operator*=(matrix &x) { return *this = *this * x; }
  matrix &operator^=(long long x) { return *this = *this ^ x; }
} dp, trans;
inline int getid(int j, int mask) { return j * 16 + mask; }
int main() {
  scanf("%d%d%d", &n, &k, &m);
  for (int j = 0; j <= k; ++j)
    for (int mask = 0; mask < (1 << m); ++mask) {
      int times = __builtin_popcount(mask) + 1;
      trans[getid(j, mask)][getid(j, (mask << 1) & (1 << m) - 1)] = 1;
      trans[getid(j, mask)][getid(j + 1, (mask << 1 | 1) & (1 << m) - 1)] =
          times;
    }
  trans = trans ^ n;
  dp[0][0] = 1;
  dp = dp * trans;
  for (int mask = 0; mask < (1 << m); ++mask) inc(ans, dp[0][getid(k, mask)]);
  printf("%d\n", ans);
  return 0;
}
