<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;
        if (goal < n)
            return 0;
        if (goal == n) {
            long long ans = 1;
            for (long long i =1 ; i<=n; ++i) {
                ans *= i;
                ans %= MOD;
            }
            return ans;
        }
        //[k][goal]
        vector<vector<long long>> dp(goal+1, vector<long long>(n+1, 0));
        dp[0][0] = 1;
        for (long long i = 1; i<=goal; ++i) {
            for (long long j = 1; j<=n; ++j) {
                dp[i][j] = 1LL * dp[i-1][j-1] * (n-j+1);
                dp[i][j] %= MOD;
                if (j > k)
                    dp[i][j] += 1LL * dp[i-1][j] * (j-k);
                dp[i][j] %= MOD;
            }
        }


        return dp[goal][n];

    }
};
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