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>
static const long long int mod = 1000000007LL;
struct matrix {
  int val[208][208];
  matrix() { memset(val, 0, sizeof(val)); }
  matrix operator*(const matrix &v) const {
    matrix ans;
    for (int i = 0; i < 208; ++i) {
      for (int j = 0; j < 208; ++j) {
        long long int a = 0;
        for (int k = 0; k < 208; ++k) {
          a += (long long int)val[i][k] * v.val[k][j] % mod;
        }
        ans.val[i][j] = a % mod;
      }
    }
    return ans;
  }
};
int n, m, k;
int cnt[16];
int main() {
  scanf("%d%d%d", &n, &k, &m);
  for (int i = 1; i < 16; ++i) cnt[i] = (cnt[i ^ (i & -i)] + 1);
  int b = (1 << m) - 1;
  matrix base;
  for (int i = 0; i <= k; ++i) {
    for (int j = 0; j <= b; ++j) {
      base.val[((j) * (k + 1) + (i))][(((j << 1) & b) * (k + 1) + (i))] += 1;
      if (i != k)
        base.val[((j) * (k + 1) + (i))]
                [((((j << 1) & b) + 1) * (k + 1) + (i + 1))] += cnt[j] + 1;
    }
  }
  matrix cur;
  for (int i = 0; i < ((k + 1) << m); ++i) cur.val[i][i] = 1;
  for (int i = 0; n != 0; ++i) {
    if ((n & (1 << i)) != 0) {
      n ^= (1 << i);
      cur = cur * base;
    }
    base = base * base;
  }
  long long int ans = 0;
  for (int i = 0; i <= b; ++i)
    ans += cur.val[((0) * (k + 1) + (0))][((i) * (k + 1) + (k))];
  printf("%d\n", (int)(ans % mod));
}
