<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;

        // dp[i][j] represents the number of playlists of length j with i unique songs
        vector<vector<long long>> dp(N + 1, vector<long long>(goal + 1, 0));

        // There's only one way to arrange zero songs (an empty playlist)
        dp[0][0] = 1;

        // Fill the dp table
        for (int i = 1; i <= N; ++i) {
            for (int j = 1; j <= goal; ++j) {
                // Case 1: We add a new unique song
                dp[i][j] = dp[i - 1][j - 1] * (N - (i - 1)) % MOD;
                
                // Case 2: We add a song that was played before (but after K songs)
                if (i > K) {
                    dp[i][j] = (dp[i][j] + dp[i][j - 1] * (i - K) % MOD) % MOD;
                }
            }
        }

        return dp[N][goal];
    }
};
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