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 maxn = 100007;
const int mod = 1e9 + 7;
long long dp[300][300];
long long ans[300][300];
int n, m, c;
int bitmask(int r) {
  int ans = 0;
  while (r) {
    if (r & 1) {
      ans++;
    }
    r >>= 1;
  }
  return ans;
}
void solve() {
  int t = n;
  for (int i = 0; i < (m + 1) * (1 << c); ++i) {
    ans[i][i] = 1;
  }
  while (t) {
    if (t & 1) {
      long long tmp[300][300];
      memset(tmp, 0, sizeof(tmp));
      for (int i = 0; i < (m + 1) * (1 << c); ++i) {
        for (int j = 0; j < (m + 1) * (1 << c); ++j) {
          for (int k = 0; k < (m + 1) * (1 << c); ++k) {
            tmp[i][j] = (tmp[i][j] + ans[i][k] * dp[k][j]) % mod;
          }
        }
      }
      for (int i = 0; i < (m + 1) * (1 << c); ++i) {
        for (int j = 0; j < (m + 1) * (1 << c); ++j) {
          ans[i][j] = tmp[i][j];
        }
      }
    }
    long long tmp[300][300];
    memset(tmp, 0, sizeof(tmp));
    for (int i = 0; i < (m + 1) * (1 << c); ++i) {
      for (int j = 0; j < (m + 1) * (1 << c); ++j) {
        for (int k = 0; k < (m + 1) * (1 << c); ++k) {
          tmp[i][j] = (tmp[i][j] + dp[i][k] * dp[k][j]) % mod;
        }
      }
    }
    for (int i = 0; i < (m + 1) * (1 << c); ++i) {
      for (int j = 0; j < (m + 1) * (1 << c); ++j) {
        dp[i][j] = tmp[i][j];
      }
    }
    t >>= 1;
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &c);
  memset(dp, 0, sizeof(dp));
  dp[0][0] = 1;
  for (int i = (1 << c); i < (m + 1) * (1 << c); ++i) {
    int r = i % (1 << c), k = i / (1 << c);
    if (r & 1 && k > 0) {
      dp[i][(k - 1) * (1 << c) + (r >> 1)] = bitmask(r);
      dp[i][(k - 1) * (1 << c) + (1 << (c - 1)) + (r >> 1)] = bitmask(r) + 1;
    } else {
      dp[i][k * (1 << c) + (r >> 1)] =
          dp[i][k * (1 << c) + (r >> 1) + (1 << (c - 1))] = 1;
    }
  }
  solve();
  long long ret = 0;
  for (int i = m * (1 << c); i < (m + 1) * (1 << c); ++i) {
    ret = (ret + ans[i][0]) % mod;
  }
  printf("%lld\n", ret);
  return 0;
}
