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 = 1000000007;
void add(int &a, long long b) { a = (a + b) % MOD; }
struct Matrix {
  vector<vector<int>> a;
  int n, m;
  Matrix(int n, int m) : n(n), m(m) { a.assign(n, vector<int>(m, 0)); }
  friend Matrix operator*(Matrix a, Matrix b) {
    Matrix c(a.n, b.m);
    for (int i = 0; i < a.n; ++i)
      for (int j = 0; j < b.m; ++j)
        for (int k = 0; k < a.m; ++k)
          add(c.a[i][j], 1LL * a.a[i][k] * b.a[k][j]);
    return c;
  }
};
Matrix identity(int n) {
  Matrix res(n, n);
  for (int i = 0; i < n; ++i) res.a[i][i] = 1;
  return res;
}
void power(Matrix &a, int n, Matrix &b) {
  while (n > 0) {
    if (n & 1) b = a * b;
    a = a * a;
    n >>= 1;
  }
}
int n, k, m;
int toId(int j, int mask) { return j * (1 << m) + mask; }
int main() {
  scanf("%d%d%d", &n, &k, &m);
  Matrix dp((k + 1) * (1 << m), 1);
  dp.a[0][0] = 1;
  Matrix f((k + 1) * (1 << m), (k + 1) * (1 << m));
  for (int j = 0; j <= k; ++j) {
    for (int mask = 0; mask < (1 << m); ++mask) {
      int newMask = (mask << 1) % (1 << m);
      int cur = toId(j, mask);
      f.a[toId(j, newMask)][cur] = 1;
      if (j < k)
        f.a[toId(j + 1, newMask | 1)][cur] = __builtin_popcount(mask) + 1;
    }
  }
  power(f, n, dp);
  int ans = 0;
  for (int mask = 0; mask < (1 << m); ++mask) add(ans, dp.a[toId(k, mask)][0]);
  printf("%d", ans);
}
