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;
struct Matrix {
  int a[20][20], n;
  void clear(int _n = 0) { n = _n, memset(a, 0, sizeof(a)); }
  Matrix(int _n = 0) { clear(_n); }
  void unit() {
    clear(n);
    for (int i = 0; i < n; i++) a[i][i] = 1;
  }
  Matrix operator*(const Matrix &m) const {
    static unsigned long long temp[20][20];
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++) temp[i][j] = 0;
    Matrix res(n);
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++)
        if (a[i][j])
          for (int k = 0; k < n; k++)
            temp[i][k] += (long long)a[i][j] * m.a[j][k];
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++) res.a[i][j] = temp[i][j] % MOD;
    return res;
  }
  Matrix operator+(const Matrix &m) const {
    Matrix res(n);
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++) res.a[i][j] = (a[i][j] + m.a[i][j]) % MOD;
    return res;
  }
};
struct Vector {
  Matrix mat[15];
  int n;
  void clear(int _n = 0, int _m = 0) {
    n = _n;
    for (int i = 0; i < n; i++) mat[i].clear(_m);
  }
  Vector(int _n = 0, int _m = 0) { clear(_n, _m); }
  Vector operator*(const Vector &v) const {
    Vector res(n, mat[0].n);
    for (int i = 0; i < n; i++)
      for (int j = 0; j <= i; j++)
        res.mat[i] = res.mat[i] + mat[j] * v.mat[i - j];
    return res;
  }
} base, res;
int main() {
  int n, K, m;
  scanf("%d%d%d", &n, &K, &m);
  base.clear(K + 1, 1 << m);
  res.clear(K + 1, 1 << m);
  res.mat[0].unit();
  for (int i = 0; i < 1 << m; i++) {
    base.mat[0].a[i][i >> 1] = 1;
    base.mat[1].a[i][i >> 1 | 1 << m >> 1] = __builtin_popcount(i) + 1;
  }
  for (int b = n; b; b >>= 1) {
    if (b & 1) res = res * base;
    base = base * base;
  }
  int ans = 0;
  for (int i = 0; i < 1 << m; i++) ans = (res.mat[K].a[0][i] + ans) % MOD;
  printf("%d\n", ans);
  return 0;
}
