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;
template <typename T>
T read1() {
  T t = 0;
  char k;
  bool vis = 0;
  do (k = getchar()) == '-' && (vis = 1);
  while ('0' > k || k > '9');
  while ('0' <= k && k <= '9')
    t = (t << 3) + (t << 1) + (k ^ '0'), k = getchar();
  return vis ? -t : t;
}
const int n = read1<long long>(), k = read1<long long>(),
          m = read1<long long>(), s = 1 << m;
const unsigned M = k + 1 << m;
struct Mat {
  int f[208][208], M;
  Mat(const unsigned n) : f() { M = n; }
  int* operator[](unsigned x) { return f[x]; }
  Mat operator*(Mat b) {
    Mat x(M);
    for (int i = 0; i < M; ++i)
      for (int j = 0; j < M; ++j)
        if (f[i][j])
          for (int k = 0; k < M; ++k)
            x[i][k] = (x[i][k] + 1ll * f[i][j] * b[j][k]) % 1000000007;
    return x;
  }
  Mat& operator*=(Mat b) { return *this = *this * b; }
  void set() {
    for (int i = 0; i < M; ++i) f[i][i] = 1;
  }
};
Mat To(M), Fi(M);
int Hash(int x, int z) { return x << m | z; }
Mat qkpow(Mat x, int n) {
  Mat t(M);
  t.set();
  for (; n; n >>= 1, x *= x)
    if (n & 1) t *= x;
  return t;
}
int main() {
  for (int i = 0; i < s; ++i)
    for (int j = 0; j <= k; ++j) {
      To[Hash(j, i)][Hash(j, (i << 1) & s - 1)] = 1;
      if (j != k)
        To[Hash(j, i)][Hash(j + 1, (i << 1 | 1) & s - 1)] =
            __builtin_popcount(i) + 1;
    }
  Fi[0][0] = 1;
  Fi *= qkpow(To, n);
  int ans = 0;
  for (int i = 0; i < s; ++i) ans = (ans + Fi[0][Hash(k, i)]) % 1000000007;
  cout << ans;
  return 0;
}
