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 maxn = 13 * 16 + 5;
const int mod = 1e9 + 7;
int n, k, m, len;
struct matrix {
  int m, n, a[maxn][maxn];
  matrix(int _m = 0, int _n = 0) {
    m = _m;
    n = _n;
    memset(a, 0, sizeof(a));
  }
  matrix operator*(const matrix &b) const {
    matrix ans(m, b.n);
    for (int i = 0; i < m; ++i)
      for (int j = 0; j < b.n; ++j)
        for (int k = 0; k < n; ++k)
          ans.a[i][j] = (ans.a[i][j] + (long long)a[i][k] * b.a[k][j]) % mod;
    return ans;
  }
} res, a;
void fpow(int k) {
  while (k) {
    if (k & 1) res = res * a;
    a = a * a;
    k >>= 1;
  }
}
int main() {
  cin >> n >> k >> m;
  res.m = res.n = a.m = a.n = len = (k + 1) * (1 << m);
  for (int i = 0; i < len; i++) res.a[i][i] = 1;
  for (int i = 0; i <= k; i++)
    for (int j = 0; j < (1 << m); j++) {
      int tp = (j << 1) % (1 << m);
      a.a[i * (1 << m) + j][i * (1 << m) + tp] = 1;
      if (i < k)
        a.a[i * (1 << m) + j][(i + 1) * (1 << m) + tp + 1] =
            __builtin_popcount(j) + 1;
    }
  fpow(n);
  memset(a.a, 0, sizeof a.a);
  a.a[0][0] = 1;
  a = a * res;
  long long ans = 0;
  for (int i = 0; i < (1 << m); i++) (ans += a.a[0][k * (1 << m) + i]) %= mod;
  cout << ans << endl;
  return 0;
}
