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 maxn = 1e5 + 10, mod = 1e9 + 7, Max1 = 13 * 5 + 10, Max2 = 14;
int n, k, m, ans = 0, ifac[20], cc[Max1][Max2];
vector<int> vec;
int Pow(int a, int b) {
  int ans = 1;
  for (; b; b >>= 1, a = 1ll * a * a % mod)
    if (b & 1) ans = 1ll * ans * a % mod;
  return ans;
}
int C(int n, int k) {
  if (n < 0 || k < 0 || n < k) return 0;
  int ans = 1;
  for (int i = 0; i < k; i++) ans = 1ll * ans * (n - i) % mod;
  return 1ll * ans * ifac[k] % mod;
}
void go(int zr = 1, int sm = n - k, int cnt = 2, int pos = 1) {
  if (sm < 0) return;
  if (pos == k) {
    ans = (1ll * ans + 1ll * zr * cc[n - sm][cnt]) % mod;
    return;
  }
  int pt = int((vec).size()) - 1, tot = 1;
  while (pt >= 0 && tot + vec[pt] <= m) {
    tot += vec[pt];
    pt--;
  }
  ++pt;
  for (int i = 0; i < m; i++) {
    vec.push_back(i + 1);
    go(1ll * zr * (int((vec).size()) - pt + 1) % mod, sm - i, cnt, pos + 1);
    vec.pop_back();
    tot++;
    while (pt < int((vec).size()) && tot > m) {
      tot -= vec[pt];
      pt++;
    }
  }
  vec.push_back(m + 1);
  go(zr, sm - m, cnt + 1, pos + 1);
  vec.pop_back();
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie();
  cin >> n >> k >> m;
  int fc = 1;
  for (int i = 0; i < 20; i++) {
    ifac[i] = Pow(fc, mod - 2);
    fc = 1ll * fc * (i + 1) % mod;
  }
  for (int i = 0; i < Max1; i++) {
    for (int j = 0; j < Max2; j++) cc[i][j] = C(n - i + j - 1, j - 1);
  }
  go();
  return cout << ans << endl, 0;
}
