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;
int n, k, m, ans;
template <typename T>
void chkmin(T &x, T y) {
  x = min(x, y);
}
template <typename T>
void chkmax(T &x, T y) {
  x = max(x, y);
}
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  x *= f;
}
int mul(const int &x, const int &y) { return 1ll * x * y % 1000000007; }
int inc(const int &x, const int &y) {
  return (x + y >= 1000000007) ? (x + y - 1000000007) : (x + y);
}
int dec(const int &x, const int &y) {
  return (x - y < 0) ? (x - y + 1000000007) : (x - y);
}
void update(int &x, const int &y) {
  x = (x + y >= 1000000007) ? (x + y - 1000000007) : (x + y);
}
int qpow(int a, int b) {
  int ret = 1;
  while (b) {
    if (b & 1) ret = mul(ret, a);
    a = mul(a, a);
    b >>= 1;
  }
  return ret;
}
int P;
struct MAT {
  int a[210][210];
} mat;
struct VEC {
  int a[210];
} vec;
MAT operator*(const MAT &a, const MAT &b) {
  MAT ret;
  for (int i = 1; i <= P; ++i)
    for (int j = 1; j <= P; ++j) {
      ret.a[i][j] = 0;
      for (int t = 1; t <= P; ++t)
        update(ret.a[i][j], mul(a.a[i][t], b.a[t][j]));
    }
  return ret;
}
VEC operator*(const MAT &a, const VEC &b) {
  VEC ret;
  for (int i = 1; i <= P; ++i) {
    ret.a[i] = 0;
    for (int t = 1; t <= P; ++t) update(ret.a[i], mul(a.a[i][t], b.a[t]));
  }
  return ret;
}
int base;
int id(int x, int y) { return x * (1 << m) + y + 1; }
int getbit(int mask) {
  int ret = 0;
  while (mask) {
    if (mask & 1) ++ret;
    mask >>= 1;
  }
  return ret;
}
int main() {
  read(n), read(k), read(m);
  base = 1 << m;
  --base;
  for (int i = 0; i <= k; ++i)
    for (int mask = 0; mask < (1 << m); ++mask) {
      int curid = id(i, mask), t = getbit(mask);
      int nmask = (mask << 1) & base, toid = id(i, nmask);
      update(mat.a[toid][curid], 1);
      if (i != k) {
        nmask |= 1, toid = id(i + 1, nmask);
        update(mat.a[toid][curid], t + 1);
      }
    }
  vec.a[id(0, 0)] = 1;
  P = id(k, base);
  MAT ret;
  for (int i = 1; i <= P; ++i)
    for (int j = 1; j <= P; ++j) ret.a[i][j] = 0;
  for (int i = 1; i <= P; ++i) ret.a[i][i] = 1;
  while (n) {
    if (n & 1) ret = ret * mat;
    mat = mat * mat;
    n >>= 1;
  }
  vec = ret * vec;
  for (int mask = 0; mask < (1 << m); ++mask) update(ans, vec.a[id(k, mask)]);
  printf("%d\n", ans);
  return 0;
}
