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;
template <typename T>
bool chkmin(T &x, T y) {
  return x > y ? x = y, 1 : 0;
}
template <typename T>
bool chkmax(T &x, T y) {
  return x < y ? x = y, 1 : 0;
}
int readint() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int cys = 1000000007;
int n, k, m, N;
long long d[215][215], g[215][215], f[215], tmp[215];
int main() {
  n = readint();
  k = readint();
  m = readint();
  N = (k + 1) * (1 << m);
  for (int i = 0; i < N; i++) {
    int t = i / (1 << m), mask = i % (1 << m);
    d[t * (1 << m) + ((mask << 1) & ((1 << m) - 1))][i]++;
    int cnt = 0;
    for (int j = 0; j < m; j++) cnt += (i >> j) & 1;
    if (t < k)
      d[(t + 1) * (1 << m) + ((mask << 1) & ((1 << m) - 1) | 1)][i] += 1 + cnt;
  }
  f[0] += 1;
  for (; n; n >>= 1) {
    if (n & 1) {
      for (int i = 0; i < N; i++)
        for (int j = 0; j < N; j++) tmp[i] = (tmp[i] + d[i][j] * f[j]) % cys;
      for (int i = 0; i < N; i++) f[i] = tmp[i], tmp[i] = 0;
    }
    for (int i = 0; i < N; i++)
      for (int j = 0; j < N; j++)
        for (int l = 0; l < N; l++)
          g[i][l] = (g[i][l] + d[i][j] * d[j][l]) % cys;
    for (int i = 0; i < N; i++)
      for (int j = 0; j < N; j++) d[i][j] = g[i][j], g[i][j] = 0;
  }
  long long ans = 0;
  for (int i = k * (1 << m); i < N; i++) ans = (ans + f[i]) % cys;
  printf("%lld\n", ans);
  return 0;
}
