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 N = 600;
const int mod = 1000000007;
int inf = 1000000000;
long long dp[2][13][40];
int g[N][N];
int n, k, m;
int tot;
int ct(int j, int mk) { return (j << m) + mk; }
void mat(int a[N][N], int b[N][N], int c[N][N]) {
  long long ret[N][N];
  for (int i = 0; i < tot; ++i)
    for (int j = 0; j < tot; ++j) ret[i][j] = 0;
  for (int k = 0; k < tot; ++k) {
    for (int i = 0; i < tot; ++i) {
      for (int j = 0; j < tot; ++j) {
        ret[i][j] = (ret[i][j] + 1ll * a[i][k] * b[k][j]) % mod;
      }
    }
  }
  for (int i = 0; i < tot; ++i) {
    for (int j = 0; j < tot; ++j) {
      c[i][j] = ret[i][j];
    }
  }
}
void pw(int a[N][N], int b, int c[N][N]) {
  int ret[N][N];
  memset(ret, 0, sizeof(ret));
  for (int i = 0; i < tot; ++i) ret[i][i] = 1;
  while (b > 0) {
    if (b & 1) mat(ret, a, ret);
    mat(a, a, a);
    b >>= 1;
  }
  for (int i = 0; i < tot; ++i)
    for (int j = 0; j < tot; ++j) c[i][j] = ret[i][j];
}
int main() {
  cin >> n >> k >> m;
  memset(dp, 0, sizeof(dp));
  long long ans = 0;
  dp[1][0][0] = 1;
  dp[1][1][1] = 1;
  for (int i = 1; i <= min(n, m); ++i) {
    int index = i & 1;
    int next_index = index ^ 1;
    memset(dp[next_index], 0, sizeof(dp[next_index]));
    for (int j = 0; j <= k; ++j) {
      int nn = min(i + 1, m);
      for (int mk = 0; mk < (1 << nn); mk++) {
        if (i == n && j == k) ans = (ans + dp[index][j][mk]) % mod;
        if (dp[index][j][mk] == 0) continue;
        int p_mk = mk;
        if (i >= m) p_mk = mk >> 1;
        int num = 0;
        for (int jj = 0; jj < nn; ++jj) {
          if ((1 << jj) & mk) ++num;
        }
        dp[next_index][j][p_mk] =
            (dp[next_index][j][p_mk] + dp[index][j][mk]) % mod;
        dp[next_index][j + 1][p_mk | (1 << (nn - 1))] =
            (dp[next_index][j + 1][p_mk | (1 << (nn - 1))] +
             dp[index][j][mk] * (num + 1)) %
            mod;
      }
    }
  }
  if (n <= m) {
    int ans = 0;
    int index = n & 1;
    for (int j = 0; j < (1 << m); ++j) {
      ans = (ans + dp[index][k][j]) % mod;
    }
    cout << ans << endl;
  } else {
    memset(g, 0, sizeof(g));
    for (int j = 0; j <= k; ++j) {
      for (int mk = 0; mk < (1 << m); mk++) {
        int p_mk = mk;
        p_mk = mk >> 1;
        int num = 0;
        for (int jj = 0; jj < m; ++jj) {
          if ((1 << jj) & mk) ++num;
        }
        g[ct(j, mk)][ct(j, p_mk)] = 1;
        g[ct(j, mk)][ct(j + 1, p_mk | (1 << (m - 1)))] = (num + 1);
      }
    }
    tot = ct(k, 1 << m);
    pw(g, n - m, g);
    long long ans = 0;
    for (int mk = 0; mk < (1 << m); ++mk) {
      for (int j = 0; j <= k; ++j) {
        for (int mmk = 0; mmk < (1 << m); ++mmk) {
          ans = (ans + dp[m & 1][j][mmk] * g[ct(j, mmk)][ct(k, mk)]) % mod;
        }
      }
    }
    cout << ans << endl;
  }
}
