<p>Your music player contains <code>n</code> different songs. You want to listen to <code>goal</code> songs (not necessarily different) during your trip. To avoid boredom, you will create a playlist so that:</p>

<ul>
	<li>Every song is played <strong>at least once</strong>.</li>
	<li>A song can only be played again only if <code>k</code> other songs have been played.</li>
</ul>

<p>Given <code>n</code>, <code>goal</code>, and <code>k</code>, return <em>the number of possible playlists that you can create</em>. Since the answer can be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>
<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 3, goal = 3, k = 1
<strong>Output:</strong> 6
<strong>Explanation:</strong> There are 6 possible playlists: [1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], and [3, 2, 1].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 2, goal = 3, k = 0
<strong>Output:</strong> 6
<strong>Explanation:</strong> There are 6 possible playlists: [1, 1, 2], [1, 2, 1], [2, 1, 1], [2, 2, 1], [2, 1, 2], and [1, 2, 2].
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 2, goal = 3, k = 1
<strong>Output:</strong> 2
<strong>Explanation:</strong> There are 2 possible playlists: [1, 2, 1] and [2, 1, 2].
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>0 &lt;= k &lt; n &lt;= goal &lt;= 100</code></li>
</ul>

class Solution:
    def numMusicPlaylists(self, n: int, goal: int, k: int) -> int:
        @cache
        def comb(a, b):
            if b == 0:
                return 0
            if b == 1:
                return a
            return comb(a, b - 1) * (a - b + 1) // b

        @cache
        def helper(n, goal, k):
            if n == 0:
                return 0
            if n - k <= 0 and goal > n:
                return 0

            ret = 1
            for i in range(goal):
                ret *= n - min(i, k)
                ret %= (10 ** 9 + 7)

            i = n - 1
            while i > 0:
                cur = comb(n, i) * helper(i, goal, k)
                cur %= (10 ** 9 + 7)
                ret += (10 ** 9 + 7)
                ret -= cur
                ret %= (10 ** 9 + 7)
                i -= 1
            return ret
                    
        return helper(n, goal, k)
18700