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 = 16 * 13;
const int Mod = 1e9 + 7;
typedef class Matrix {
 public:
  int r, c;
  int a[S][S];
  Matrix() {}
  Matrix(int r, int c) : r(r), c(c) {}
  Matrix operator*(Matrix b) {
    Matrix rt(r, b.c);
    for (int i = 0; i < r; i++) {
      for (int j = 0; j < b.c; j++) {
        rt[i][j] = 0;
      }
    }
    for (int i = 0; i < r; i++) {
      for (int j = 0; j < c; j++) {
        for (int k = 0; k < b.c; k++) {
          rt[i][k] = (rt[i][k] + 1ll * a[i][j] * b[j][k]) % Mod;
        }
      }
    }
    return rt;
  }
  int* operator[](int p) { return a[p]; }
} Matrix;
int n, m, k;
int bit[16];
Matrix F0, T, pT;
int main() {
  scanf("%d%d%d", &n, &k, &m);
  int s = (k + 1) << m;
  F0.r = 1, F0.c = s;
  F0[0][0] = 1;
  T.r = s, T.c = s;
  int msk_and = (1 << m) - 1;
  for (int i = 1; i <= msk_and; i++) bit[i] = bit[i >> 1] + (i & 1);
  for (int c = 0; c <= k; c++) {
    for (int msk = 0; msk <= msk_and; msk++) {
      int nmsk = (msk << 1) & msk_and;
      T[(c << m) | msk][(c << m) | nmsk]++;
      if (c < k) {
        T[(c << m) | msk][((c + 1) << m) | nmsk | 1] += bit[msk] + 1;
      }
    }
  }
  pT.r = s, pT.c = s;
  for (int i = 0; i < s; i++) {
    pT[i][i] = 1;
  }
  for (; n; n >>= 1, T = T * T) {
    if (n & 1) {
      pT = pT * T;
    }
  }
  F0 = F0 * pT;
  int ans = 0;
  for (int msk = 0; msk <= msk_and; msk++) {
    ans += F0[0][(k << m) | msk];
    (ans >= Mod) && (ans -= Mod);
  }
  printf("%d\n", ans);
  return 0;
}
