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;
using ll = long long;
const int P = 1000000007;
int n, m, kk;
struct Matrix {
  vector<vector<int>> a;
  int n, m;
  Matrix(int n, int m) : n(n), m(m) { a.assign(n, vector<int>(m, 0)); }
};
Matrix operator*(const Matrix &a, const Matrix &b) {
  Matrix c(a.n, b.m);
  for (int i = 0; i < a.n; i++)
    for (int j = 0; j < b.m; j++)
      for (int k = 0; k < a.m; k++)
        c.a[i][j] = (c.a[i][j] + a.a[i][k] * 1ll * b.a[k][j]) % P;
  return c;
}
void qpow(Matrix a, int b, Matrix &ret) {
  while (b) {
    if (b & 1) ret = ret * a;
    a = a * a;
    b >>= 1;
  }
}
inline int ID(int i, int msk) { return i * (1 << m) + msk; }
int main() {
  cin >> n >> kk >> m;
  Matrix dp(1, (kk + 1) * (1 << m)),
      f((kk + 1) * (1 << m), (kk + 1) * (1 << m));
  dp.a[0][0] = 1;
  for (int i = 0; i <= kk; i++)
    for (int msk = 0; msk < (1 << m); msk++) {
      int nm = (msk << 1) & ((1 << m) - 1), cur = ID(i, msk);
      f.a[cur][ID(i, nm)] = 1;
      if (i < kk) f.a[cur][ID(i + 1, nm | 1)] = __builtin_popcount(msk) + 1;
    }
  qpow(f, n, dp);
  ll ans = 0;
  for (int msk = 0; msk < (1 << m); msk++)
    ans = (ans + dp.a[0][ID(kk, msk)]) % P;
  cout << ans << endl;
  return 0;
}
