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 maxn = 100007;
const int inf = (1LL << 29);
const int mod = 1e9 + 7;
struct matrix {
  int n, m;
  long long shu[210][210];
  matrix() { memset(shu, 0, sizeof(shu)); }
  matrix operator*(matrix a) {
    matrix new1;
    new1.n = n;
    new1.m = a.m;
    for (int i = 1; i <= n; i++)
      for (int k = 1; k <= a.m; k++)
        for (int z = 1; z <= m; z++)
          new1.shu[i][k] += shu[i][z] * a.shu[z][k], new1.shu[i][k] %= mod;
    return new1;
  }
  matrix operator+(matrix a) {
    matrix new1;
    new1.n = n;
    new1.m = m;
    for (int i = 1; i <= n; i++)
      for (int k = 1; k <= m; k++) {
        new1.shu[i][k] = (shu[i][k] + a.shu[i][k]) % mod;
      }
    return new1;
  }
  void print() {
    for (int i = 1; i <= n; i++) {
      for (int k = 1; k <= m; k++) {
        printf("%d ", shu[i][k]);
      }
      printf("\n");
    }
  }
  void eye() {
    for (int i = 1; i <= n; i++) shu[i][i] = 1;
  }
};
matrix powmat(matrix a, int b) {
  matrix ans;
  int first = 0;
  while (b) {
    if (b & 1) {
      if (!first)
        ans = a;
      else
        ans = ans * a;
      first = 1;
    }
    a = a * a;
    b >>= 1;
  }
  return ans;
}
int n, tot, m;
int id(int x, int y) { return x * (1 << m) + y + 1; }
matrix a, b, c;
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  cin >> n >> tot >> m;
  a.shu[1][id(0, 0)] = 1;
  int x = tot * (1 << m) + (1 << m);
  a.n = a.m = b.n = b.m = x;
  for (int k = 0; k <= tot; k++) {
    for (int z = 0; z < (1 << m); z++) {
      int nex = (z << 1) % (1 << m);
      b.shu[id(k, z)][id(k, nex)] = 1;
      if (k < tot) {
        int x = __builtin_popcount(z) + 1;
        b.shu[id(k, z)][id(k + 1, nex | 1)] = x;
      }
    }
  }
  a = a * powmat(b, n);
  int ans = 0;
  for (int i = 0; i < (1 << m); i++) {
    (ans += a.shu[1][id(tot, i)]) %= mod;
  }
  cout << ans;
  return 0;
}
