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;
using ld = long double;
vector<vector<vector<ll>>> v;
vector<vector<ll>> mat;
vector<vector<ll>> matMal(vector<vector<ll>> mat1, vector<vector<ll>> mat2) {
  vector<vector<ll>> mat3((int)(mat1).size(), vector<ll>((int)(mat1).size()));
  for (int i = 0; i < (int)(mat1).size(); ++i) {
    for (int j = 0; j < (int)(mat1).size(); ++j) {
      ll sum = 0;
      for (int k = 0; k < (int)(mat1).size(); ++k) {
        sum = sum + (ll)mat1[i][k] * mat2[k][j] % 1000000007;
      }
      mat3[i][j] = sum % 1000000007;
    }
  }
  return mat3;
}
vector<vector<ll>> matMult(vector<vector<ll>> matr, ll x) {
  if (x == 1) return matr;
  if (x % 2 == 1) return matMal(matMult(matr, x - 1), mat);
  vector<vector<ll>> matTemp = matMult(matr, x / 2);
  return matMal(matTemp, matTemp);
}
int main() {
  int n, k, m;
  vector<ll> anzb = {0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4};
  cin >> n >> k >> m;
  mat.resize((1 << m) * (k + 1), vector<ll>((1 << m) * (k + 1)));
  for (int i = 0; i < 1 << m; ++i) {
    for (int j = 0; j < k + 1; ++j) {
      if (i % 2 == 0) {
        mat[(k + 1) * i + j][(k + 1) * (i / 2) + j] = 1;
        mat[(k + 1) * i + j][(k + 1) * ((i + (1 << m)) / 2) + j] = 1;
      } else if (j != 0) {
        mat[(k + 1) * i + j][(k + 1) * (i / 2) + j - 1] = anzb[i / 2] + 1;
        mat[(k + 1) * i + j][(k + 1) * ((i + (1 << m)) / 2) + j - 1] =
            anzb[(i + (1 << m)) / 2] + 1;
      }
    }
  }
  mat = matMult(mat, n);
  ll sum = 0;
  for (int i = 0; i < 1 << m; ++i) {
    sum = (sum + mat[(k + 1) * i + k][0]) % 1000000007;
  }
  cout << sum << endl;
}
