<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 N;
    int Goal;
    int K;
    int mod=1e9+7;
    long f(int countsong,int countunique,vector<vector<long>>&dp)
    {
        if(countsong==Goal)
        {
            if(countunique==N)
            return 1;

            else
            return 0;
        }
        long result=0;

        if(dp[countsong][countunique]!=-1)
        return dp[countsong][countunique];

        if(N-countunique>0) //if we are taking unique elements the number of options I have is the number unique elemets I have taken subtracted by number of elemets left
        result+=(N-countunique)*f(countsong+1,countunique+1,dp)%mod;

        if(countunique-K>0) //if we are taking elements that we have already taken we need atleast k unique elements in between this element and previsly when it was taken  
        
        result+=(countunique-K)*f(countsong+1,countunique,dp)%mod;

        return dp[countsong][countunique]=result%mod;

    }
    int numMusicPlaylists(int n, int goal, int k) {

        N=n;
        Goal=goal;
        K=k;

        vector<vector<long>> dp(goal+1,vector<long>(goal+1,-1));

        return f(0,0,dp);
       
    }
};
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