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 N = 222;
const int MO = 1e9 + 7;
int mul(int x, int y) { return (long long)x * y % MO; }
struct node {
  int a[N][N];
} b, c;
int n;
long long w[N][N];
node operator*(node p, node q) {
  node r;
  int i, j, k;
  for (i = 0; i < n; i = i + 1)
    for (j = 0; j < n; j = j + 1)
      for (k = 0; k < n; k = k + 1) w[i][k] += mul(p.a[i][j], q.a[j][k]);
  for (i = 0; i < n; i = i + 1)
    for (j = 0; j < n; j = j + 1) r.a[i][j] = w[i][j] % MO, w[i][j] = 0;
  return r;
}
int l, k, m;
int main() {
  int i, j, x, y, z;
  long long s;
  cin >> l >> k >> m;
  n = (k + 1) << m;
  for (i = 0; i < n; i = i + 1) {
    x = i >> m;
    y = i - (x << m);
    z = 0;
    for (j = 0; j < m; j = j + 1)
      if (y & (1 << j)) z++;
    y >>= 1;
    b.a[i][x << m | y] = 1;
    if (x == k) continue;
    x++;
    y |= 1 << (m - 1);
    b.a[i][x << m | y] = z + 1;
  }
  for (i = 0; i < n; i = i + 1) c.a[i][i] = 1;
  while (l) {
    if (l & 1) c = c * b;
    b = b * b;
    l >>= 1;
  }
  s = 0;
  for (i = 0; i < (1 << m); i = i + 1) s += c.a[0][k << m | i];
  cout << s % MO;
  return 0;
}
