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;
int n, K, m;
int T[13][16][13][16];
long long f[210][210];
int Num(int x) {
  int ans = 0;
  while (x != 0) {
    ans += x % 2;
    x /= 2;
  }
  return ans;
}
int Js(int x, int y) { return (x << m) + y; }
long long c[210][210];
void Chen(long long a[][210], long long b[][210]) {
  memset(c, 0, sizeof(c));
  for (int i = 0; i < ((K + 1) << m); i++)
    for (int j = 0; j < ((K + 1) << m); j++)
      for (int k = 0; k < ((K + 1) << m); k++)
        c[i][j] = (c[i][j] + a[i][k] * b[k][j]) % mod;
  for (int i = 0; i < ((K + 1) << m); i++)
    for (int j = 0; j < ((K + 1) << m); j++) a[i][j] = c[i][j];
}
long long ans[210][210];
void Pow(long long a[][210], int b) {
  for (int i = 0; i < ((K + 1) << m); i++) ans[i][i] = 1;
  while (b != 0) {
    if (b % 2) Chen(ans, a);
    Chen(a, a);
    b /= 2;
  }
}
long long anss;
int main() {
  scanf("%d %d %d", &n, &K, &m);
  for (int i = 1; i <= K; i++)
    for (int j = 0; j < (1 << m); j++) {
      int num = Num(j);
      if (num > i) continue;
      if (j % 2) {
        T[i - 1][j / 2][i][j] = num;
        T[i - 1][j / 2 + (1 << (m - 1))][i][j] = num + 1;
      } else {
        T[i][j / 2][i][j] = 1;
        T[i][j / 2 + (1 << (m - 1))][i][j] = 1;
      }
    }
  T[0][0][0][0] = 1;
  for (int i = 0; i <= K; i++)
    for (int j = 0; j < (1 << m); j++)
      for (int k = 0; k <= K; k++)
        for (int r = 0; r < (1 << m); r++) {
          f[Js(i, j)][Js(k, r)] = T[i][j][k][r];
        }
  Pow(f, n);
  for (int i = 0; i < ((K + 1) << m); i++)
    if (i >= Js(K, 0)) {
      anss = (anss + ans[0][i]) % mod;
    }
  printf("%d", anss);
  return 0;
}
