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 maxc = 210;
const int mod = 1e9 + 7;
struct matrix {
  int x[maxc][maxc];
  matrix() { memset(x, 0, sizeof(x)); }
};
matrix dp, mul;
int nod, n, k, m, id[15][20];
matrix operator*(matrix A, matrix B) {
  matrix C;
  for (int i = 1; i <= nod; ++i)
    for (int j = 1; j <= nod; ++j)
      for (int t = 1; t <= nod; ++t)
        C.x[i][j] = (C.x[i][j] + 1LL * A.x[i][t] * B.x[t][j]) % mod;
  return C;
}
matrix operator^(matrix A, int B) {
  matrix C;
  for (int i = 1; i <= nod; ++i) C.x[i][i] = 1;
  for (; B; A = A * A, B >>= 1)
    if (B & 1) C = C * A;
  return C;
}
void readf() {
  cin >> n >> k >> m;
  nod = 0;
  for (int i = 0; i <= k; ++i)
    for (int mask = 0; mask < (1 << m); ++mask) id[i][mask] = ++nod;
  dp.x[1][id[0][0]] = dp.x[1][id[1][1]] = 1;
}
void solve() {
  for (int i = 0; i <= k; ++i)
    for (int mask = 0; mask < (1 << m); ++mask) {
      int newmas = (mask << 1) & ((1 << m) - 1);
      mul.x[id[i][mask]][id[i][newmas]] = 1;
      mul.x[id[i][mask]][id[i + 1][newmas | 1]] =
          __builtin_popcount((mask)) + 1;
    }
  if (n > 0) dp = dp * (mul ^ (n - 1));
  int ans = 0;
  for (int mask = 0; mask < (1 << m); ++mask)
    ans = (ans + dp.x[1][id[k][mask]]) % mod;
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  readf();
  solve();
}
