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;
namespace RA {
int rnd(int p) { return 1ll * rand() * rand() % p; }
int rnd(int L, int R) { return rnd(R - L + 1) + L; }
}  // namespace RA
const int N = 1e5 + 5, K = 13, S = 1 << 5, P = 1e9 + 7;
void add(int& x, long long y) { x = (x + y) % P; }
template <const int SZ>
struct Matrix {
  int n, m, c[SZ][SZ];
  Matrix() {}
  Matrix(int _n, int _m) {
    assert(0 <= _n && _n < SZ);
    assert(0 <= _m && _m < SZ);
    n = _n, m = _m;
    for (int i = (1); i <= (n); ++i)
      for (int j = (1); j <= (m); ++j) c[i][j] = 0;
  }
  int* operator[](int x) {
    assert(0 <= x && x < SZ);
    return c[x];
  }
  Matrix operator*(Matrix M) {
    assert(m == M.n);
    Matrix res(n, M.m);
    for (int i = (1); i <= (n); ++i)
      for (int j = (1); j <= (m); ++j)
        if (c[i][j])
          for (int k = (1); k <= (M.m); ++k) {
            add(res[i][k], 1ll * c[i][j] * M[j][k]);
          }
    return res;
  }
};
Matrix<K * S + 1> mtx(K*S - 1, K*S - 1), a(K*S - 1, K*S - 1);
int n, k, m, lim;
int _(int j, int s) { return j * lim + s + 1; }
void addTrans(int j1, int s1, int j2, int s2, int coef) {
  add(mtx[_(j1, s1)][_(j2, s2)], coef);
}
int f[K][S], g[K][S];
int main() {
  scanf("%d%d%d", &n, &k, &m);
  f[0][0] = 1;
  lim = 1 << m;
  for (int j = (0); j <= (k); ++j)
    for (int s = (0); s <= (lim - 1); ++s) {
      addTrans(j, s, j, (s << 1) & (lim - 1), 1);
      if (j < k) {
        addTrans(j, s, j + 1, (s << 1 | 1) & (lim - 1),
                 __builtin_popcount(s) + 1);
      }
    }
  for (int i = (1); i <= (K * S - 1); ++i) a[i][i] = 1;
  while (n) {
    if (n & 1) a = a * mtx;
    mtx = mtx * mtx;
    n >>= 1;
  }
  int ans = 0;
  for (int s = (0); s <= (lim - 1); ++s) add(ans, a[_(0, 0)][_(k, s)]);
  printf("%d\n", ans);
  return 0;
}
