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 readint() {
  int a = 0;
  char c = getchar(), f = 1;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') f = -f;
  for (; '0' <= c && c <= '9'; c = getchar())
    a = (a << 3) + (a << 1) + (c ^ 48);
  return a * f;
}
const int Mod = 1e9 + 7;
const int MaxN = (1 << 4) * 13;
struct Matrix {
  int a[MaxN][MaxN];
  Matrix operator*(const Matrix &b) const {
    Matrix c;
    memset(c.a, 0, MaxN * MaxN << 2);
    for (int i = (0); i <= (MaxN - 1); ++i)
      for (int j = (0); j <= (MaxN - 1); ++j)
        if (a[i][j])
          for (int k = (0); k <= (MaxN - 1); ++k)
            c.a[i][k] = (1ll * a[i][j] * b.a[j][k] + c.a[i][k]) % Mod;
    return c;
  }
};
inline Matrix qkpow(Matrix b, int q) {
  Matrix a = b;
  --q;
  for (; q; q >>= 1, b = b * b)
    if (q & 1) a = a * b;
  return a;
}
Matrix Q;
int n, m, k, cnt[1 << 4];
void solve() {
  for (int i = 1; i < (1 << m); ++i) cnt[i] = cnt[i ^ (i & -i)] + 1;
  for (int S = 0; S < (1 << m); ++S)
    for (int i = 0; i < k; ++i) {
      int S_ = (S << 1 | 1) & ((1 << m) - 1);
      if (i + 1 < k) Q.a[S * k + i][S_ * k + i + 1] += cnt[S] + 1;
      ++Q.a[S * k + i][(S_ ^ 1) * k + i];
    }
  Q = qkpow(Q, n);
  ;
  int ans = 0;
  for (int i = 0; i < MaxN; ++i)
    if (i % k == k - 1) ans = (ans + Q.a[0][i]) % Mod;
  printf("%d\n", ans);
}
int main() {
  n = readint();
  k = readint() + 1;
  m = readint();
  solve();
  return 0;
}
