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 S = 13, M = 4, SZ = S * (1 << M), P = 1e9 + 7;
int n, K, m, s, sum;
struct Mat {
  int w[SZ][SZ], n, m;
  Mat inline operator*(Mat b) {
    Mat c;
    memset(c.w, 0, sizeof c.w);
    c.n = n, c.m = b.m;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < b.m; j++) {
        for (int k = 0; k < m; k++) {
          c.w[i][j] = (c.w[i][j] + (long long)w[i][k] * b.w[k][j]) % P;
        }
      }
    }
    return c;
  }
} A, res;
int inline get(int i, int j) { return i * (1 << m) + j; }
int inline bits(int x) {
  int cnt = 0;
  while (x) cnt++, x -= x & -x;
  return cnt;
}
void inline add(int u, int v, int w) { (A.w[u][v] += w) %= P; }
int inline An(int x) { return x & s; }
int main() {
  scanf("%d%d%d", &n, &K, &m);
  s = (1 << m) - 1;
  sum = (K + 1) * (1 << m);
  A.n = A.m = res.m = sum, res.n = 1;
  res.w[0][0] = 1;
  for (int i = 0; i <= K; i++) {
    for (int j = 0; j < (1 << m); j++) {
      add(get(i, j), get(i, An(j << 1)), 1);
      if (i < K) add(get(i, j), get(i + 1, An(j << 1 | 1)), 1 + bits(j));
    }
  }
  while (n) {
    if (n & 1) res = res * A;
    A = A * A;
    n >>= 1;
  }
  int ans = 0;
  for (int i = 0; i < (1 << m); i++) ans = (ans + res.w[0][get(K, i)]) % P;
  printf("%d\n", ans);
  return 0;
}
