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;
using ll = long long;
const int inf = 0x3f3f3f3f;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
struct Matrix {
  ll m[210][210];
  int n;
  Matrix(int _n = 0, int _v = 0) {
    n = _n;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++) m[i][j] = (i == j) ? _v : 0;
  }
  Matrix operator*(const Matrix& o) const {
    Matrix ret(n, 0);
    for (int k = 0; k < n; k++)
      for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++) (ret.m[i][j] += m[i][k] * o.m[k][j]) %= mod;
    return ret;
  }
  Matrix operator^(int k) const {
    Matrix ret(n, 1), A = *this;
    while (k) {
      if (k & 1) ret = ret * A;
      A = A * A;
      k >>= 1;
    }
    return ret;
  }
  void show() {
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++) cout << m[i][j] << " \n"[j == n - 1];
  }
};
int n, k, m, N;
int getCol(int i, int s) { return i * (N + 1) + s; }
int main() {
  cin >> n >> k >> m;
  N = (1 << m) - 1;
  Matrix A((k + 1) * (N + 1), 0);
  for (int i = 0; i <= k; i++)
    for (int s = 0; s <= N; s++) {
      int ns = (s << 1) & N;
      int c = getCol(i, s);
      A.m[c][getCol(i, ns)] += 1;
      if (i < k) {
        A.m[c][getCol(i + 1, ns | 1)] += __builtin_popcount(s) + 1;
      }
    }
  A = A ^ n;
  ll ans = 0;
  for (int i = 0; i <= N; i++) ans += A.m[0][getCol(k, i)];
  cout << ans % mod << endl;
  return 0;
}
