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 mod = 1e9 + 7;
void add(long long &a, long long b) { a = (a + b) % mod; }
long long f[2][1005][1024];
int pc[1024];
long long solve(int n, int k, int m) {
  memset(f, 0, sizeof f);
  f[0][0][0] = 1;
  for (int i = 0; i < n; ++i) {
    int x = i & 1, y = 1 - x;
    memset(f[y], 0, sizeof f[y]);
    for (int j = 0; j <= k; ++j) {
      for (int mask = 0; mask < (1 << m); ++mask) {
        int new_mask = mask >> 1;
        add(f[y][j][new_mask], f[x][j][mask]);
        add(f[y][j + 1][new_mask + (1 << m - 1)],
            f[x][j][mask] * (pc[mask] + 1));
      }
    }
  }
  long long ret = 0;
  for (int mask = 0; mask < (1 << m); ++mask) add(ret, f[n & 1][k][mask]);
  return ret;
}
vector<vector<long long>> operator*(const vector<vector<long long>> &a,
                                    const vector<vector<long long>> &b) {
  vector<vector<long long>> c(a.size(), vector<long long>(b[0].size(), 0));
  for (int i = 0; i < a.size(); ++i)
    for (int j = 0; j < b.size(); ++j)
      for (int k = 0; k < b[0].size(); ++k) add(c[i][k], a[i][j] * b[j][k]);
  return c;
}
vector<vector<long long>> operator+(const vector<vector<long long>> &a,
                                    const vector<vector<long long>> &b) {
  vector<vector<long long>> c = a;
  for (int i = 0; i < a.size(); ++i)
    for (int j = 0; j < a[0].size(); ++j) add(c[i][j], b[i][j]);
  return c;
}
vector<vector<long long>> A(int m) {
  vector<vector<long long>> a(1 << m, vector<long long>(1 << m, 0));
  for (int i = 0; i < (1 << m - 1); ++i) a[i][i * 2] = a[i][i * 2 + 1] = 1;
  return a;
}
vector<vector<long long>> B(int m) {
  vector<vector<long long>> b(1 << m, vector<long long>(1 << m, 0));
  for (int i = 0; i < (1 << m - 1); ++i) {
    b[i + (1 << m - 1)][i * 2] = pc[i] + 1;
    b[i + (1 << m - 1)][i * 2 + 1] = pc[i] + 2;
  }
  return b;
}
vector<vector<long long>> U(int m) {
  vector<vector<long long>> u(1 << m, vector<long long>(1 << m, 0));
  for (int i = 0; i < (1 << m); ++i) u[i][i] = 1;
  return u;
}
vector<vector<long long>> O(int m) {
  return vector<vector<long long>>(1 << m, vector<long long>(1 << m, 0));
}
void convo(vector<vector<vector<long long>>> &a,
           vector<vector<vector<long long>>> b, int m, int k) {
  int sz = min(k + 1, (int)(a.size() + b.size() + 1));
  a.resize(sz, O(m));
  for (int i = sz - 1; i >= 0; --i) {
    a[i] = a[i] * b[0];
    for (int j = 1; j <= i && j < b.size(); ++j) a[i] = a[i] + a[i - j] * b[j];
  }
}
long long solve3(int n, int k, int m) {
  vector<vector<vector<long long>>> ans, mul;
  --n;
  ans = mul = {A(m), B(m)};
  while (n) {
    if (n & 1) convo(ans, mul, m, k);
    convo(mul, mul, m, k);
    n >>= 1;
  }
  long long ret = 0;
  for (int i = 0; i < (1 << m); ++i) add(ret, ans[k][i][0]);
  return ret;
}
int main(void) {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k, m;
  for (int i = 1; i < 1024; ++i) pc[i] = pc[i - (i & -i)] + 1;
  while (cin >> n >> k >> m) cout << solve3(n, k, m) << endl;
  return 0;
}
