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;
int n, k, m;
int S, M;
int bit[250];
struct mat {
  int a[250][250];
  mat(int d = 0) {
    memset(a, 0, sizeof a);
    for (int i = 0; i < 250; i++) a[i][i] = d;
  }
  mat operator*(const mat &B) const {
    mat ret = mat(0);
    for (int i = 0; i < S; i++)
      for (int j = 0; j < S; j++)
        if (a[i][j])
          for (int k = 0; k < S; k++)
            if (B.a[j][k])
              ret.a[i][k] =
                  (ret.a[i][k] + a[i][j] * 1ll * B.a[j][k]) % 1000000007;
    return ret;
  }
} A;
mat Pow(mat base, int k) {
  mat ret = mat(1);
  for (; k; k >>= 1, base = base * base)
    if (k & 1) ret = ret * base;
  return ret;
}
int main() {
  scanf("%d%d%d", &n, &k, &m);
  M = 1 << m;
  S = (k + 1) * M;
  for (int sta = 1; sta < M; sta++) bit[sta] = bit[sta - (sta & -sta)] + 1;
  for (int sta = 0; sta < M; sta++)
    for (int i = 0; i <= k; i++) {
      A.a[i * M + sta][i * M + ((sta << 1) & (M - 1))]++;
      if (i < k)
        A.a[i * M + sta][(i + 1) * M + ((sta << 1) & (M - 1)) + 1] +=
            bit[sta] + 1;
    }
  A = Pow(A, n);
  int ans = 0;
  for (int sta = 0; sta < M; sta++) {
    ans = (ans + A.a[0][k * M + sta]) % 1000000007;
  }
  printf("%d\n", (ans + 1000000007) % 1000000007);
}
