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 std::cin;
using std::cout;
const int mod = 1e9 + 7;
const int N = 13 << 4;
struct Mat {
  int a[N][N];
  inline Mat operator*(const Mat& b) const {
    Mat ret;
    memset(&ret, 0, sizeof ret);
    for (int i = 0; i < N; ++i)
      for (int j = 0; j < N; ++j)
        if (a[i][j]) {
          for (int k = 0; k < N; ++k) {
            ret.a[i][k] = (ret.a[i][k] + (long long)a[i][j] * b.a[j][k]) % mod;
          }
        }
    return ret;
  }
} base, trans, I;
inline Mat pow(Mat a, int b) {
  Mat ans = I;
  for (; b; b >>= 1, a = a * a) {
    if (b & 1) {
      ans = ans * a;
    }
  }
  return ans;
}
int n, k, m;
inline int idx(int cnt, int s) { return cnt << m | s; }
int main() {
  std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  for (int i = 0; i < N; ++i) I.a[i][i] = 1;
  cin >> n >> k >> m;
  for (int i = 0; i <= k; ++i) {
    for (int j = 0; j < 1 << m; ++j) {
      trans.a[idx(i, j)][idx(i, j >> 1)] = 1;
      if (i != k) {
        trans.a[idx(i, j)][idx(i + 1, j >> 1 | 1 << m - 1)] =
            __builtin_popcount(j) + 1;
      }
    }
  }
  base.a[0][0] = 1;
  trans = base * pow(trans, n);
  long long ans = 0;
  for (int j = 0; j < 1 << m; ++j) {
    ans += trans.a[0][idx(k, j)];
  }
  cout << ans % mod << '\n';
}
