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 MAX_N = 100001;
const long long MOD = 1000000007;
const long long INF = 1e18;
using mat = array<array<long long, 224>, 224>;
long long add(long long a, long long b) { return (a + b) % MOD; }
long long sub(long long a, long long b) { return (a - b + MOD) % MOD; }
long long mul(long long a, long long b) { return (a * b) % MOD; }
long long ADD(long long &a, long long b) { return a = add(a, b); }
long long SUB(long long &a, long long b) { return a = sub(a, b); }
long long MUL(long long &a, long long b) { return a = mul(a, b); }
long long n, m, k;
mat trans, x;
mat mul(const mat &a, const mat &b) {
  for (int i = (0); i < ((k + 1) << m); i++) x[i].fill(0ll);
  for (int i = (0); i < ((k + 1) << m); i++)
    for (int l = (0); l < ((k + 1) << m); l++)
      for (int j = (0); j < ((k + 1) << m); j++)
        ADD(x[i][j], mul(a[i][l], b[l][j]));
  return x;
}
mat po(const mat &a, int b) {
  if (b == 1) return a;
  return ((b & 1) ? (mul(a, po(a, b - 1))) : (po(mul(a, a), b / 2)));
}
int main(int argc, const char *argv[]) {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> n >> k >> m;
  for (int i = (0); i < (k << m); i++) trans[i].fill(0ll);
  for (int j = (0); j < (k + 1); j++)
    for (int mask = (0); mask < (1 << m); mask++) {
      trans[(j << m) + mask][(j << m) + (((1 << m) - 1) & (mask << 1))]++;
      long long num = 1;
      for (int l = (0); l < (m); l++) num += (mask >> l) & 1;
      if (j != k)
        trans[(j << m) + mask]
             [((j + 1) << m) + (((1 << m) - 1) & (mask << 1)) + 1] += num;
    }
  mat dp = po(trans, n);
  long long ans = 0;
  for (int mask = (0); mask < (1 << m); mask++)
    ADD(ans, dp[0][(k << m) + mask]);
  cout << ans << endl;
  return 0;
}
