This problem is same as the next one, but has smaller 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 ≤ 10^5, 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>
long long n, k, m;
long long dp[100005][14][1 << 4];
signed main() {
  scanf("%lld %lld %lld", &n, &k, &m);
  dp[0][0][0] = 1;
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j <= k; j++)
      for (long long s = 0; s < (1 << m); s++) {
        dp[i + 1][j][s >> 1] =
            (dp[i + 1][j][s >> 1] + dp[i][j][s]) % 1000000007;
        dp[i + 1][j + 1][s >> 1 | 1 << m - 1] =
            (dp[i + 1][j + 1][s >> 1 | 1 << m - 1] +
             dp[i][j][s] * (__builtin_popcount(s) + 1) % 1000000007) %
            1000000007;
      }
  long long ans = 0;
  for (long long i = 0; i < (1 << m); i++)
    ans = (ans + dp[n][k][i]) % 1000000007;
  printf("%lld\n", ans);
  return 0;
}
