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 = 1000000007;
int mul(long long a, long long b) { return a * b % mod; }
using Matrix = vector<vector<int>>;
Matrix create(int n, int m) {
  return vector<vector<int>>(n, vector<int>(m, 0));
}
Matrix iden(int n) {
  Matrix res = create(n, n);
  for (int i = 0; i < n; i++) res[i][i] = 1;
  return res;
}
Matrix mul(Matrix &a, Matrix &b) {
  int n = a.size();
  Matrix res = create(n, n);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      for (int k = 0; k < n; k++) {
        res[i][j] += mul(a[i][k], b[k][j]);
        if (res[i][j] >= mod) res[i][j] -= mod;
      }
  return res;
}
int main() {
  int n, k, m;
  cin >> n >> k >> m;
  int len = ((k + 1) << m) + 1;
  Matrix upd = create(len, len);
  for (int i = 0; i < k << m; i++) {
    int hm = i & ((1 << m) - 1);
    int hk = i >> m;
    int nm = ((hm << 1) + 1) & ((1 << m) - 1);
    upd[i][((hk + 1) << m) + nm] = 1 + __builtin_popcount(hm);
    nm &= ~1;
    upd[i][(hk << m) + nm] = 1;
  }
  for (int i = k << m; i < len; i++) upd[i][len - 1] = 1;
  Matrix res = iden(len);
  for (; n; n >>= 1, upd = mul(upd, upd))
    if (n & 1) res = mul(res, upd);
  int ans = 0;
  for (int i = k << m; i < len; i++) {
    ans += res[0][i];
    if (ans >= mod) ans -= mod;
  }
  cout << ans << endl;
  return 0;
}
