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 = 1e9 + 7;
inline int add(int a, int b) { return (a + b >= MOD) ? a + b - MOD : a + b; }
inline void inc(int& a, int b) { a = add(a, b); }
inline int sub(int a, int b) { return (a - b < 0) ? a - b + MOD : a - b; }
inline void dec(int& a, int b) { a = sub(a, b); }
inline int mul(int a, int b) { return 1ll * a * b % MOD; }
inline void grow(int& a, int b) { a = mul(a, b); }
struct Mat {
  int n, m;
  vector<vector<int>> mat;
  Mat() {}
  Mat(int _n, int _m) {
    n = _n, m = _m;
    mat.assign(n, vector<int>(m, 0));
  }
  void id() {
    assert(n == m);
    for (int i = 0; i < n; i++) mat[i][i] = 1;
  }
  Mat operator*(const Mat& o) const {
    assert(m == o.n);
    Mat res = Mat(n, o.m);
    for (int i = 0; i < n; i++)
      for (int j = 0; j < o.m; j++)
        for (int k = 0; k < m; k++)
          inc(res.mat[i][j], mul(mat[i][k], o.mat[k][j]));
    return res;
  }
};
Mat bpow(Mat a, int b) {
  assert(a.n == a.m);
  Mat res(a.n, a.n);
  res.id();
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
int n, m, k, siz, ans;
Mat res, trans;
int hsh(int i, int j) { return i * (1 << m) + j; }
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin >> n >> k >> m;
  siz = (k + 1) * (1 << m);
  res = Mat(siz, 1);
  res.mat[hsh(0, 0)][0] = 1;
  trans = Mat(siz, siz);
  for (int i = 0; i <= k; i++) {
    for (int j = 0; j < (1 << m); j++) {
      if (j & 1) {
        if (!i) continue;
        inc(trans.mat[hsh(i, j)][hsh(i - 1, j >> 1)],
            1 + __builtin_popcount(j >> 1));
        inc(trans.mat[hsh(i, j)][hsh(i - 1, (j >> 1) ^ (1 << (m - 1)))],
            2 + __builtin_popcount(j >> 1));
      } else {
        inc(trans.mat[hsh(i, j)][hsh(i, j >> 1)], 1);
        inc(trans.mat[hsh(i, j)][hsh(i, (j >> 1) ^ (1 << (m - 1)))], 1);
      }
    }
  }
  res = bpow(trans, n) * res;
  for (int i = 0; i < (1 << m); i++) inc(ans, res.mat[hsh(k, i)][0]);
  cout << ans << endl;
}
