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 long long mod = 1e9 + 7;
long long n, k, m;
struct matrix {
  long long s[230][230];
  matrix() { memset(s, 0, sizeof(s)); }
} cs, zy;
long long count(long long x) {
  long long sum = 0;
  while (x) x -= (x & -x), sum++;
  return sum;
}
long long getid(long long x, long long y) { return x * (1 << m) + y; }
matrix operator*(matrix a, matrix b) {
  matrix c;
  for (long long i = 0; i < (k + 1) * (1 << m); i++) {
    for (long long j = 0; j < (k + 1) * (1 << m); j++) {
      for (long long l = 0; l < (k + 1) * (1 << m); l++) {
        c.s[i][j] = (c.s[i][j] + a.s[i][l] * b.s[l][j]) % mod;
      }
    }
  }
  return c;
}
matrix operator^(matrix a, long long b) {
  matrix c;
  for (long long i = 0; i < (k + 1) * (1 << m); i++) c.s[i][i] = 1;
  while (b) {
    if (b & 1) c = c * a;
    a = a * a;
    b >>= 1;
  }
  return c;
}
void solve() {
  cs.s[0][0] = 1;
  for (long long i = 0; i <= k; i++) {
    for (long long j = 0; j < (1 << m); j++) {
      zy.s[getid(i, j)][getid(i, j >> 1)] = 1;
      zy.s[getid(i, j)][getid(i + 1, (j >> 1) | (1 << (m - 1)))] = 1 + count(j);
    }
  }
  cs = cs * (zy ^ n);
}
signed main() {
  scanf("%lld%lld%lld", &n, &k, &m);
  solve();
  long long ans = 0;
  for (long long i = 0; i < (1 << m); i++)
    ans = (ans + cs.s[0][getid(k, i)]) % mod;
  printf("%lld\n", ans);
  return 0;
}
