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 = 212, MOD = 1e9 + 7;
long long tmp[maxN][maxN];
void mult(long long one[maxN][maxN], long long two[maxN][maxN],
          long long resp[maxN][maxN]) {
  memset(tmp, 0, sizeof tmp);
  for (int i = 0; i < maxN; i++)
    for (int j = 0; j < maxN; j++)
      for (int k = 0; k < maxN; k++)
        tmp[i][j] = (tmp[i][j] + one[i][k] * two[k][j]) % MOD;
  memcpy(resp, tmp, sizeof tmp);
}
long long ott[maxN][maxN];
void fastPow(long long one[maxN][maxN], long long b,
             long long resp[maxN][maxN]) {
  memset(ott, 0, sizeof ott);
  for (int i = 0; i < maxN; i++) ott[i][i] = 1;
  while (b) {
    if (b & 1) mult(ott, one, ott);
    b >>= 1;
    mult(one, one, one);
  }
  memcpy(resp, ott, sizeof ott);
}
int n, m, k;
long long base[maxN][maxN], strt[maxN][maxN];
void consider(int to, int fr, int x, int cnd) {
  for (int i = 0; i + cnd <= k; i++) {
    base[(i + cnd) * (1 << m) + to][i * (1 << m) + fr] = x;
  }
}
int coun(int x) {
  int ret = 0;
  while (x) {
    if (x & 1) ret++;
    x >>= 1;
  }
  return ret;
}
int main() {
  cin >> n >> k >> m;
  for (int i = 0; i < (1 << m); i++) {
    int cr = i >> 1;
    consider(cr, i, 1, 0);
    cr |= 1 << (m - 1);
    consider(cr, i, coun(i) + 1, 1);
  }
  strt[0][0] = 1;
  fastPow(base, n, base);
  mult(base, strt, base);
  long long ans = 0;
  for (int i = 0; i < (1 << m); i++)
    ans = (ans + base[k * (1 << m) + i][0]) % MOD;
  cout << ans << '\n';
}
