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;
struct Mar {
  long long a[225][225];
  Mar() { memset(a, 0, sizeof(a)); }
  Mar operator*(const Mar B) {
    Mar c;
    for (int i = 0; i < 225; i++) {
      for (int j = 0; j < 225; j++) {
        for (int k = 0; k < 225; k++) {
          c.a[i][j] = (c.a[i][j] + a[i][k] * B.a[k][j]) % mod;
        }
      }
    }
    return c;
  }
};
void pw(Mar &g, Mar f, long long b) {
  while (b) {
    if (b & 1) g = f * g;
    b >>= 1;
    f = f * f;
  }
}
long long n, sz[20], S;
int get(int k, int s) { return k * S + s; }
int main() {
  int k, m;
  scanf("%lld%d%d", &n, &k, &m);
  for (int i = 1; i < (1 << m); i++) {
    sz[i] = sz[i >> 1] + (i & 1);
  }
  S = (1 << m);
  Mar f, g;
  g.a[0][0] = 1;
  for (int i = 0; i <= k; i++) {
    for (int s = 0; s < S; s++) {
      int ns = (s << 1) & (S - 1);
      f.a[get(i, ns)][get(i, s)] = 1;
      f.a[get(i + 1, ns | 1)][get(i, s)] = sz[s] + 1;
    }
  }
  pw(g, f, n);
  long long ans = 0;
  for (int s = 0; s < S; s++) {
    ans = (ans + g.a[get(k, s)][0]) % mod;
  }
  printf("%lld", ans);
  return 0;
}
