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 M = 4;
const int K = 13;
const int N = K * (1 << M) + 1;
const int MOD = 1000000007;
inline int add(int u, int v) { return (u += v) >= MOD ? u - MOD : u; }
inline int sub(int u, int v) { return (u -= v) < 0 ? u + MOD : u; }
inline int mul(int u, int v) { return (long long)u * v % MOD; }
struct Matrix {
  int a[N][N];
  Matrix(int diag = 0) {
    memset(a, 0, sizeof a);
    for (int i = 0; i < N; i++) a[i][i] = diag;
  }
  int* operator[](int u) { return a[u]; }
  Matrix operator*(Matrix& u) const {
    Matrix res;
    for (int k = 0; k < N; k++) {
      for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
          res[i][j] = add(res[i][j], mul(a[i][k], u[k][j]));
        }
      }
    }
    return res;
  }
  Matrix operator^(int v) const {
    Matrix res(1);
    Matrix u = *this;
    while (v) {
      if (v & 1) res = res * u;
      u = u * u;
      v >>= 1;
    }
    return res;
  }
  void debug(int n) {
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) cout << a[i][j];
      cout << endl;
    }
    cout << endl;
  }
};
int n, m, k;
int encode(int u, int mask) { return (u << m) + mask; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k >> m;
  Matrix b;
  for (int i = 0; i <= k; i++) {
    for (int mask = 0; mask < (1 << m); mask++) {
      int id = encode(i, mask);
      int id2 = encode(i, (mask << 1) & ((1 << m) - 1));
      b[id][id2] = 1;
      if (i < k) {
        id2 = encode(i + 1, (mask << 1) & ((1 << m) - 1) | 1);
        b[id][id2] = __builtin_popcount(mask) + 1;
      }
    }
  }
  Matrix a;
  a[0][0] = 1;
  Matrix c = b ^ n;
  c = a * c;
  int res = 0;
  for (int mask = 0; mask < (1 << m); mask++) {
    res = add(res, c[0][encode(k, mask)]);
  }
  cout << res << endl;
  return 0;
}
