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 long long mod = 1e9 + 7;
struct SquareMatrix {
  SquareMatrix(int _n) : n(_n) { data.assign(_n, vector<long long>(_n, 0)); }
  vector<long long> &operator[](int i) { return data[i]; }
  const vector<long long> &operator[](int i) const { return data[i]; }
  SquareMatrix operator*(const SquareMatrix &other) const {
    assert(n == other.n);
    SquareMatrix ret(n);
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        for (int k = 0; k < n; ++k) {
          ret[i][j] = (ret[i][j] + data[i][k] * other[k][j]) % mod;
        }
      }
    }
    return ret;
  }
  vector<vector<long long>> data;
  int n;
};
SquareMatrix quickpower(SquareMatrix m, long long p) {
  int n = m.n;
  SquareMatrix ret(n);
  for (int i = 0; i < n; ++i) ret[i][i] = 1;
  while (p) {
    if (p & 1) {
      ret = ret * m;
    }
    m = m * m;
    p >>= 1;
  }
  return ret;
}
int fun(int x) {
  int res = 0;
  while (x) {
    res += x % 2;
    x /= 2;
  }
  return res;
}
int main() {
  long long n, k, m;
  scanf("%lld %lld %lld", &n, &k, &m);
  long long num = 1 << m;
  SquareMatrix M(num * (k + 1));
  for (int i = 0; i < num * (k + 1); i++) {
    int x = i % (k + 1);
    int y = i / (k + 1);
    int x1 = x;
    int y1 = (y * 2) % num;
    int x2 = x + 1;
    int y2 = (y * 2) % num + 1;
    M[i][y1 * (k + 1) + x1] += 1;
    if (x2 <= k) M[i][y2 * (k + 1) + x2] += fun(y) + 1ll;
  }
  M = quickpower(M, n);
  long long ans = 0;
  for (int i = 0; i < num; i++) ans = (ans + M[0][i * (k + 1) + k]) % mod;
  cout << ans << endl;
  return 0;
}
