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>
void mult(int aa[][((1 << 4) * (12 + 1) * 2)],
          int bb[][((1 << 4) * (12 + 1) * 2)],
          int cc[][((1 << 4) * (12 + 1) * 2)], int n) {
  int i, j, k;
  for (i = 0; i < n; i++) memset(cc[i], 0, n * sizeof *cc[i]);
  for (i = 0; i < n; i++)
    for (k = 0; k < n; k++) {
      int *bk = bb[k], a = aa[i][k];
      if (a)
        for (j = 0; j < n; j++) {
          int b = bk[j];
          if (b) cc[i][j] = (cc[i][j] + (long long)a * b) % 1000000007;
        }
    }
}
void power(int aa[][((1 << 4) * (12 + 1) * 2)],
           int pp[][((1 << 4) * (12 + 1) * 2)],
           int tt[][((1 << 4) * (12 + 1) * 2)], int n, int k) {
  if (k == 0) {
    int i, j;
    for (i = 0; i < n; i++)
      for (j = 0; j < n; j++) pp[i][j] = i == j;
    return;
  }
  if (k % 2 == 0) {
    power(aa, tt, pp, n, k / 2);
    mult(tt, tt, pp, n);
  } else {
    power(aa, pp, tt, n, k / 2);
    mult(pp, pp, tt, n);
    mult(tt, aa, pp, n);
  }
}
int main() {
  static int cnt[1 << 4],
      aa[((1 << 4) * (12 + 1) * 2)][((1 << 4) * (12 + 1) * 2)],
      pp[((1 << 4) * (12 + 1) * 2)][((1 << 4) * (12 + 1) * 2)],
      tt[((1 << 4) * (12 + 1) * 2)][((1 << 4) * (12 + 1) * 2)];
  int n, k, m, b, c, h, head, ans;
  scanf("%d%d%d", &n, &k, &m);
  for (b = 1; b < 1 << m; b++) cnt[b] = cnt[b & b - 1] + 1;
  for (b = 0; b < 1 << m; b++)
    for (c = 0; c <= k; c++)
      for (head = 0; head < 2; head++) {
        int i, j, b_, c_, head_;
        i = (b * (k + 1) + c) * 2 + head;
        b_ = (b & (1 << m - 1) - 1) << 1, c_ = c, head_ = b >> m - 1;
        if (!head || !head_) {
          j = (b_ * (k + 1) + c_) * 2 + (head || head_);
          aa[i][j] = 1;
        }
        if (c < k) {
          c_ = c + 1;
          if (!head || !head_) {
            j = ((b_ | 1) * (k + 1) + c_) * 2 + (head || head_);
            aa[i][j] = 1;
            j = (b_ * (k + 1) + c_) * 2 + (head || head_);
            aa[i][j] = cnt[b] + head;
          }
          for (h = 0; h < m; h++)
            if (b & 1 << h) {
              b_ = ((b ^ 1 << h) & (1 << m - 1) - 1) << 1,
              head_ = (b ^ 1 << h) >> m - 1;
              if (!head || !head_) {
                j = ((b_ | 1) * (k + 1) + c_) * 2 + (head || head_);
                aa[i][j] = 1;
                j = (b_ * (k + 1) + c_) * 2 + (head || head_);
                aa[i][j] = cnt[b] + head - 1;
              }
            }
        }
      }
  power(aa, pp, tt, (1 << m) * (k + 1) * 2, n);
  c = k;
  ans = 0;
  for (b = 0; b < 1 << m; b++)
    for (head = 0; head < 2; head++)
      if (head == 0 && cnt[b] == 1 || head == 1 && b == 0) {
        int i = (b * (k + 1) + c) * 2 + head;
        ans = (ans + pp[0][i]) % 1000000007;
      }
  printf("%d\n", ans);
  return 0;
}
