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;
template <typename T>
void maxtt(T& t1, T t2) {
  t1 = max(t1, t2);
}
template <typename T>
void mintt(T& t1, T t2) {
  t1 = min(t1, t2);
}
bool debug = 0;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
string direc = "URDL";
long long ln, lk, lm;
void etp(bool f = 0) {
  puts(f ? "YES" : "NO");
  exit(0);
}
void addmod(int& x, int y, int mod = 1000000007) {
  assert(y >= 0);
  x += y;
  if (x >= mod) x -= mod;
  assert(x >= 0 && x < mod);
}
void et(int x = -1) {
  printf("%d\n", x);
  exit(0);
}
long long fastPow(long long x, long long y, int mod = 1000000007) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1) ans = (x * ans) % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
long long gcd1(long long x, long long y) { return y ? gcd1(y, x % y) : x; }
struct Matrix {
  int n, m;
  int a[215][215];
  Matrix() {}
  Matrix(int n, int m) : n(n), m(m) {}
  void init(int num) {
    for (int(i) = 0; (i) < (int)(n); (i)++)
      for (int(j) = 0; (j) < (int)(m); (j)++) a[i][j] = num;
  }
  Matrix operator*(Matrix& b) {
    Matrix o = Matrix(n, b.m);
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < b.m; j++) {
        o.a[i][j] = 0;
        for (int k = 0; k < m; k++) {
          addmod(o.a[i][j], (long long)a[i][k] * b.a[k][j] % 1000000007);
        }
      }
    }
    return o;
  }
};
int ans;
int id(int k, int msk) { return k * (1 << m) + msk; }
void fmain(int tid) {
  scanf("%d%d%d", &n, &k, &m);
  int N = (k + 1) * (1 << m);
  Matrix mt = Matrix(N, N);
  mt.init(0);
  for (int j = 0; j <= k; j++)
    for (int msk = 0; msk < (1 << m); msk++) {
      int fm = id(j, msk);
      mt.a[fm][id(j, msk >> 1)]++;
      if (j < k) {
        int to = id(j + 1, (msk >> 1) | (1 << (m - 1)));
        int cc = __builtin_popcount(msk);
        mt.a[fm][to] = cc + 1;
      }
    }
  Matrix b = Matrix(1, N);
  b.init(0);
  b.a[0][id(0, 0)] = 1;
  while (n > 0) {
    if (n & 1) b = b * mt;
    mt = mt * mt;
    n >>= 1;
  }
  for (int(msk) = 0; (msk) < (int)(1 << m); (msk)++)
    addmod(ans, b.a[0][id(k, msk)]);
  printf("%d\n", ans);
}
int main() {
  int t = 1;
  for (int(i) = 1; (i) <= (int)(t); (i)++) {
    fmain(i);
  }
  return 0;
}
