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>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
using namespace std;
const int MN = 210, MK = 13, MM = (1 << 4), inf = 1000000005, mod = 1000000007;
const long long INF = 1000000000000000005LL;
struct matrix {
  long long V[MN][MN];
  matrix(long long prz = 1LL) {
    for (int i = 0; i < MN; ++i)
      for (int j = 0; j < MN; ++j) V[i][j] = 0LL;
    for (int i = 0; i < MN; ++i) V[i][i] = prz;
  }
  matrix operator*(matrix& m) {
    matrix wyn(0LL);
    for (int i = 0; i < MN; ++i)
      for (int j = 0; j < MN; ++j)
        for (int k = 0; k < MN; ++k) {
          wyn.V[i][j] += V[i][k] * m.V[k][j];
          if (wyn.V[i][j] >= mod) wyn.V[i][j] %= mod;
        }
    return wyn;
  }
  matrix operator^(long long wyk) {
    matrix wyn(1LL), tmp = (*this);
    while (wyk) {
      if (wyk & 1LL) wyn = wyn * tmp;
      tmp = tmp * tmp;
      wyk >>= 1;
    }
    return wyn;
  }
};
int main() {
  int n, k, m;
  scanf("%d%d%d", &n, &k, &m);
  matrix A(0LL);
  matrix B(0LL);
  A.V[0][0] = 1LL;
  int M = (1 << m), K = k + 1;
  for (int j = 0; j <= k; ++j)
    for (int mask = 0; mask < M; ++mask) {
      int a = mask * K + j;
      int m1, m2;
      m1 = (mask >> 1);
      m2 = m1 + (1 << (m - 1));
      if (mask & 1) {
        if (j > 0) {
          int b = m1 * K + j - 1;
          B.V[b][a] += (1 + __builtin_popcount(m1));
          b = m2 * K + j - 1;
          B.V[b][a] += (1 + __builtin_popcount(m2));
        }
      } else {
        int b = m1 * K + j;
        B.V[b][a]++;
        b = m2 * K + j;
        B.V[b][a]++;
      }
    }
  B = B ^ n;
  A = A * B;
  long long suma = 0LL;
  for (int i = 0; i < M; ++i) suma += A.V[0][i * K + k];
  printf("%lld", suma % mod);
}
