<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 {
public:
       int numMusicPlaylists(int n, int goal, int k) {
        const int MOD = 1e9 + 7;
        vector dp(goal + 1, vector<long>(n + 1, -1));

        function<long(int, int)> numberOfPlaylists = [&](int i, int j) -> long {
            // Base cases
            if (i == 0 && j == 0) {
                return 1LL;
            }
            if (i == 0 || j == 0) {
                return 0LL;
            }
            if (dp[i][j] != -1) {
                return dp[i][j];
            }
            // DP transition: add a new song or replay an old one
            dp[i][j] = (numberOfPlaylists(i - 1, j - 1) * (n - j + 1)) % MOD;
            if (j > k) {
                dp[i][j] += (numberOfPlaylists(i - 1, j) * (j - k)) % MOD;
                dp[i][j] %= MOD;
            }
            return dp[i][j];
        };

        return numberOfPlaylists(goal, n);
    }
};
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