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 = 100005;
const int mod = 1e9 + 7;
int n, k, m, all;
inline void add(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
struct mat {
  int a[210][210];
  mat() { memset(a, 0, sizeof a); }
  inline mat operator*(const mat &o) const {
    mat re;
    for (int k = 0; k < all; ++k)
      for (int i = 0; i < all; ++i)
        if (a[i][k])
          for (int j = 0; j < all; ++j)
            if (o.a[k][j]) add(re.a[i][j], 1ll * a[i][k] * o.a[k][j] % mod);
    return re;
  }
  inline mat operator^(int b) const {
    mat re, A = *this;
    for (int i = 0; i < all; ++i) re.a[i][i] = 1;
    while (b) {
      if (b & 1) re = re * A;
      A = A * A;
      b >>= 1;
    }
    return re;
  }
};
inline int enc(int K, int S) { return K * (1 << m) + S; }
inline int nxt(int S, bool x) { return ((S << 1) | x) & ((1 << m) - 1); }
int cnt[16];
int main() {
  scanf("%d%d%d", &n, &k, &m);
  all = (k + 1) * (1 << m);
  mat trans, ans;
  ans.a[0][enc(0, 0)] = 1;
  for (int s = 1; s < (1 << m); ++s) cnt[s] = cnt[s >> 1] + (s & 1);
  for (int i = 0; i <= k; ++i)
    for (int s = 0; s < (1 << m); ++s) {
      if (i < k) trans.a[enc(i, s)][enc(i + 1, nxt(s, 1))] = cnt[s] + 1;
      trans.a[enc(i, s)][enc(i, nxt(s, 0))] = 1;
    }
  ans = ans * (trans ^ n);
  int Ans = 0;
  for (int s = 0; s < (1 << m); ++s) add(Ans, ans.a[0][enc(k, s)]);
  printf("%d\n", (Ans + mod) % mod);
}
