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;
inline int read() {
  char c = getchar();
  int x = 0;
  bool f = 0;
  for (; !isdigit(c); c = getchar()) f ^= !(c ^ 45);
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  if (f) x = -x;
  return x;
}
struct modint {
  int x;
  modint(int o = 0) { x = o; }
  modint &operator=(int o) { return x = o, *this; }
  modint &operator+=(modint o) {
    return x = x + o.x >= 1000000007 ? x + o.x - 1000000007 : x + o.x, *this;
  }
  modint &operator-=(modint o) {
    return x = x - o.x < 0 ? x - o.x + 1000000007 : x - o.x, *this;
  }
  modint &operator*=(modint o) { return x = 1ll * x * o.x % 1000000007, *this; }
  modint &operator^=(int b) {
    modint a = *this, c = 1;
    for (; b; b >>= 1, a *= a)
      if (b & 1) c *= a;
    return x = c.x, *this;
  }
  modint &operator/=(modint o) { return *this *= o ^= 1000000007 - 2; }
  modint &operator+=(int o) {
    return x = x + o >= 1000000007 ? x + o - 1000000007 : x + o, *this;
  }
  modint &operator-=(int o) {
    return x = x - o < 0 ? x - o + 1000000007 : x - o, *this;
  }
  modint &operator*=(int o) { return x = 1ll * x * o % 1000000007, *this; }
  modint &operator/=(int o) { return *this *= ((modint(o)) ^= 1000000007 - 2); }
  template <class I>
  friend modint operator+(modint a, I b) {
    return a += b;
  }
  template <class I>
  friend modint operator-(modint a, I b) {
    return a -= b;
  }
  template <class I>
  friend modint operator*(modint a, I b) {
    return a *= b;
  }
  template <class I>
  friend modint operator/(modint a, I b) {
    return a /= b;
  }
  friend modint operator^(modint a, int b) { return a ^= b; }
  friend bool operator==(modint a, int b) { return a.x == b; }
  friend bool operator!=(modint a, int b) { return a.x != b; }
  bool operator!() { return !x; }
  modint operator-() { return x ? 1000000007 - x : 0; }
};
int sz;
struct mat {
  modint a[305][305];
  void init() { memset(a, 0, sizeof a); }
  mat() { init(); }
  void init1() {
    for (register int i = (0); i <= (sz); ++i)
      for (register int j = (0); j <= (sz); ++j) a[i][j] = (i == j);
  }
  inline modint *operator[](const int &x) { return a[x]; }
  mat operator*(const mat &b) const {
    mat c;
    for (register int i = (0); i <= (sz); ++i)
      for (register int k = (0); k <= (sz); ++k)
        for (register int j = (0); j <= (sz); ++j)
          c.a[i][j] += a[i][k] * b.a[k][j];
    return c;
  }
};
mat t;
int n, m, k, cnt[1 << 4], mxs;
inline int P(int x, int y) { return x * mxs + y; }
inline mat qpow(mat a, int b) {
  mat res;
  res.init1();
  for (; b; b >>= 1, a = a * a)
    if (b & 1) res = res * a;
  return res;
}
signed main() {
  n = read(), k = read(), m = read();
  mxs = (1 << m);
  sz = P(k, mxs - 1);
  for (register int i = (1); i <= (mxs - 1); ++i)
    cnt[i] = cnt[i >> 1] + (i & 1);
  for (register int j = (0); j <= (k); ++j) {
    for (register int s = (0); s <= (mxs - 1); ++s) {
      int s1 = (s >> 1);
      int s2 = (s >> 1) | (1 << (m - 1));
      if ((s & 1) == 0) {
        t[P(j, s)][P(j, s1)] += 1;
        t[P(j, s)][P(j, s2)] += 1;
        continue;
      }
      if (j > 0) {
        t[P(j, s)][P(j - 1, s1)] += (1 + cnt[s1]);
        t[P(j, s)][P(j - 1, s2)] += (1 + cnt[s2]);
      }
    }
  }
  t = qpow(t, n);
  modint res = 0;
  for (register int i = (0); i <= (mxs); ++i) res += t[P(k, i)][P(0, 0)];
  cout << res.x;
  return 0;
}
