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 MXN = (int)3e2 + 7;
const int mod = (int)1e9 + 7;
int add(int v1, int v2) {
  v1 += v2;
  if (v1 >= mod) v1 -= mod;
  return v1;
}
struct Mat {
  int n, m;
  long long a[MXN][MXN];
  void init(int _n, int _m) {
    n = _n, m = _m;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++) {
        a[i][j] = 0;
      }
  }
  Mat operator*(const Mat &p2) {
    Mat res;
    res.init(n, p2.m);
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++)
        for (int k = 0; k < p2.m; k++) {
          res.a[i][k] = (res.a[i][k] + a[i][j] * p2.a[j][k]) % mod;
        }
    return res;
  }
  Mat operator^(const long long &p2) {
    long long t = p2 - 1;
    Mat res = *this, x = *this;
    while (t) {
      if (t & 1) res = res * x;
      t >>= 1;
      x = x * x;
    }
    return res;
  }
} a;
int main() {
  int n, k, m;
  scanf("%d %d %d", &n, &k, &m);
  a.init((k + 1) * (1 << m), (k + 1) * (1 << m));
  for (int j = 0; j <= k; j++) {
    for (int mask = 0; mask < (1 << m); mask++) {
      int nxt = (mask << 1) % (1 << m);
      int v1 = j * (1 << m) + mask;
      int v2 = j * (1 << m) + nxt;
      a.a[v2][v1] = 1;
      if (j != k) {
        int num = __builtin_popcount(mask) + 1;
        v2 = (j + 1) * (1 << m) + nxt + 1;
        a.a[v2][v1] = num;
      }
    }
  }
  a = a ^ n;
  int ans = 0;
  for (int mask = 0; mask < (1 << m); mask++) {
    int v = k * (1 << m) + mask;
    ans = add(ans, a.a[v][0]);
  }
  printf("%d\n", ans);
  return 0;
}
