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>
const int K = 14;
const int M = 1 << 4;
const int MX = 310;
const long long MOD = 1e9 + 7;
using namespace std;
int n, m, k;
int tot = 0, maxn;
int g[K][M];
void read(int &x) {
  char c;
  for (; c > '9' || c < '0'; c = getchar())
    ;
  for (; '0' <= c && c <= '9'; c = getchar())
    x = (x << 3) + (x << 1) + (c ^ 48);
}
struct nn {
  long long a[MX][MX];
  void MEM() {
    for (int i = 0; i <= maxn; i++)
      for (int j = 0; j <= maxn; j++) a[i][j] = 0;
  }
} hlp;
nn MUL(nn A, nn B) {
  nn C;
  C.MEM();
  for (int i = 1; i <= tot; i++)
    for (int j = 1; j <= tot; j++) {
      C.a[i][j] = 0;
      for (int r = 1; r <= tot; r++) {
        C.a[i][j] += A.a[i][r] * B.a[r][j] % MOD;
        C.a[i][j] %= MOD;
      }
    }
  return C;
}
long long KSM(int y) {
  nn res;
  res.MEM();
  long long ans;
  res.a[1][1] = 1;
  while (y) {
    if (y % 2 == 1) res = MUL(res, hlp);
    hlp = MUL(hlp, hlp);
    y /= 2;
  }
  for (int num = 0; num <= maxn; num++) ans = (ans + res.a[1][g[k][num]]) % MOD;
  return ans;
}
int main() {
  read(n);
  read(k);
  read(m);
  maxn = (1 << m) - 1;
  for (int i = 0; i <= k; i++)
    for (int j = 0; j <= maxn; j++) g[i][j] = ++tot;
  for (int len = 0; len <= k; len++)
    for (int num = 0; num <= maxn; num++) {
      int num1 = (num << 1) & maxn;
      hlp.a[g[len][num]][g[len][num1]] = 1;
      if (len < k)
        hlp.a[g[len][num]][g[len + 1][num1 | 1]] = __builtin_popcount(num) + 1;
    }
  printf("%lld\n", KSM(n));
  return 0;
}
