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;
struct matrix {
  int rows, cols, MOD;
  int **c;
  void init() {
    c = new int *[rows];
    for (int i = 0; i < rows; i++) c[i] = new int[cols];
  }
  void clean() {
    for (int i = 0; i < rows; i++) delete[] c[i];
    delete[] c;
  }
  matrix(int row, int col, int M = 1000000007) : rows(row), cols(col), MOD(M) {
    init();
    for (int i = 0; i < rows; i++) memset(c[i], 0, cols * sizeof(int));
  }
  matrix(const matrix &o) : rows(o.rows), cols(o.cols), MOD(o.MOD) {
    init();
    for (int i = 0; i < rows; i++) memcpy(c[i], o.c[i], cols * sizeof(int));
  }
  matrix &operator=(const matrix &o) {
    clean();
    rows = o.rows;
    cols = o.cols;
    MOD = o.MOD;
    init();
    for (int i = 0; i < rows; i++) memcpy(c[i], o.c[i], cols * sizeof(int));
    return *this;
  }
  ~matrix() { clean(); }
  void print() {
    for (int i = 0; i < rows; i++) {
      for (int j = 0; j < cols; j++) printf("%d ", c[i][j]);
      puts("");
    }
  }
  matrix operator+(matrix b) const {
    assert(rows == b.rows && cols == b.cols);
    for (int i = 0; i < rows; i++)
      for (int j = 0; j < cols; j++) {
        b.c[i][j] += c[i][j];
        if (b.c[i][j] >= MOD) b.c[i][j] -= MOD;
      }
    return b;
  }
  matrix operator-(matrix b) const {
    assert(rows == b.rows && cols == b.cols);
    matrix ret = *this;
    for (int i = 0; i < rows; i++)
      for (int j = 0; j < cols; j++) {
        ret.c[i][j] -= b.c[i][j];
        if (ret.c[i][j] < 0) ret.c[i][j] += MOD;
      }
    return ret;
  }
  matrix operator*(long long k) const {
    matrix ret = *this;
    k %= MOD;
    for (int i = 0; i < rows; i++)
      for (int j = 0; j < cols; j++)
        ret.c[i][j] = (1LL * ret.c[i][j] * k) % MOD;
    return ret;
  }
  matrix operator%(matrix b) const {
    assert(cols == b.rows);
    matrix ret(rows, b.cols, MOD);
    long long MOD2 = 1LL * MOD * MOD;
    for (int i = 0; i < rows; i++) {
      for (int j = 0; j < b.cols; j++) {
        long long sum = 0;
        for (int k = 0; k < cols; k++) {
          sum += (1LL * c[i][k] * b.c[k][j]);
          if (sum >= MOD2) sum -= MOD2;
        }
        ret.c[i][j] = sum % MOD;
      }
    }
    return ret;
  }
  matrix operator^(long long exp) const {
    matrix ret = matrix(rows, cols, MOD), base = *this;
    for (int i = 0; i < rows; i++) ret.c[i][i] = 1;
    while (exp) {
      if (exp & 1) ret = ret % base;
      exp >>= 1;
      if (exp) base = base % base;
    }
    return ret;
  }
};
int n, k, m, dim;
int t(int m, int r) { return m * (k + 1) + r; }
int main() {
  scanf("%d %d %d", &n, &k, &m);
  dim = (k + 1) * (1 << m);
  matrix mt(dim, dim);
  for (int msk = 0; msk < (1 << m); msk++) {
    mt.c[t(msk, 0)][t(msk, 0)] = 1;
    for (int rem = 1; rem <= k; rem++) {
      int nmsk = ((msk << 1) & ((1 << m) - 1));
      mt.c[t(msk, rem)][t(nmsk, rem)] = 1;
      mt.c[t(msk, rem)][t(nmsk | 1, rem - 1)] = (1LL + __builtin_popcount(msk));
    }
  }
  mt = (mt ^ n);
  long long ans = 0;
  for (int i = 0; i < (1 << m); i++) {
    ans += mt.c[t(0, k)][t(i, 0)];
    if (ans >= mt.MOD) ans -= mt.MOD;
  }
  printf("%lld\n", ans);
  return 0;
}
