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;
vector<vector<int>> mul(const vector<vector<int>>& A,
                        const vector<vector<int>>& B) {
  if (A[0].size() != B.size()) {
    throw out_of_range("Nu poti face inmultire");
  }
  vector<vector<int>> C(A.size(), vector<int>(B[0].size()));
  for (int i = 0; i < (int)C.size(); ++i)
    for (int k = 0; k < (int)A[0].size(); ++k)
      for (int j = 0; j < (int)C[0].size(); ++j)
        C[i][j] = (C[i][j] + 1LL * A[i][k] * B[k][j]) % mod;
  return C;
}
vector<vector<int>> Identity(int n) {
  vector<vector<int>> M(n, vector<int>(n));
  for (int i = 0; i < n; ++i) M[i][i] = 1;
  return M;
}
int main() {
  int n, k, m;
  cin >> n >> k >> m;
  vector<vector<int>> dp(k + 1, vector<int>(1 << m));
  vector<vector<int>> ndp = dp;
  vector<int> nb(1 << m);
  for (int i = 1; i < (1 << m); ++i) nb[i] = nb[i - (i & -i)] + 1;
  auto getMatrix = [&]() {
    int stari = (1 << m) * (k + 1);
    vector<vector<int>> mat(stari, vector<int>(stari));
    vector<vector<int>> id(k + 1, vector<int>(1 << m));
    int NEXT_FREE_ID = 0;
    for (int iau = 0; iau <= k; ++iau)
      for (int mask = 0; mask < 1 << m; ++mask) id[iau][mask] = NEXT_FREE_ID++;
    for (int iau = 0; iau <= k; ++iau) {
      for (int mask = 0; mask < 1 << m; ++mask) {
        int nmask = mask;
        if (nmask & (1 << (m - 1))) {
          nmask -= 1 << (m - 1);
        }
        nmask <<= 1;
        mat[id[iau][mask]][id[iau][nmask]]++;
        if (iau + 1 <= k) {
          nmask++;
          mat[id[iau][mask]][id[iau + 1][nmask]] += nb[mask] + 1;
        }
      }
    }
    return mat;
  };
  vector<vector<int>> raise = getMatrix();
  vector<vector<int>> curr = Identity(raise.size());
  for (int y = n; y > 0; y /= 2) {
    if (y & 1) {
      curr = mul(curr, raise);
    }
    raise = mul(raise, raise);
  }
  int res = 0;
  for (int i = 0; i < (1 << m); ++i)
    res = (res + curr[0][k * (1 << m) + i]) % mod;
  cout << res;
}
