<p>A concert hall has <code>n</code> rows numbered from <code>0</code> to <code>n - 1</code>, each with <code>m</code> seats, numbered from <code>0</code> to <code>m - 1</code>. You need to design a ticketing system that can allocate seats in the following cases:</p>

<ul>
	<li>If a group of <code>k</code> spectators can sit <strong>together</strong> in a row.</li>
	<li>If <strong>every</strong> member of a group of <code>k</code> spectators can get a seat. They may or <strong>may not</strong> sit together.</li>
</ul>

<p>Note that the spectators are very picky. Hence:</p>

<ul>
	<li>They will book seats only if each member of their group can get a seat with row number <strong>less than or equal</strong> to <code>maxRow</code>. <code>maxRow</code> can <strong>vary</strong> from group to group.</li>
	<li>In case there are multiple rows to choose from, the row with the <strong>smallest</strong> number is chosen. If there are multiple seats to choose in the same row, the seat with the <strong>smallest</strong> number is chosen.</li>
</ul>

<p>Implement the <code>BookMyShow</code> class:</p>

<ul>
	<li><code>BookMyShow(int n, int m)</code> Initializes the object with <code>n</code> as number of rows and <code>m</code> as number of seats per row.</li>
	<li><code>int[] gather(int k, int maxRow)</code> Returns an array of length <code>2</code> denoting the row and seat number (respectively) of the <strong>first seat</strong> being allocated to the <code>k</code> members of the group, who must sit <strong>together</strong>. In other words, it returns the smallest possible <code>r</code> and <code>c</code> such that all <code>[c, c + k - 1]</code> seats are valid and empty in row <code>r</code>, and <code>r &lt;= maxRow</code>. Returns <code>[]</code> in case it is <strong>not possible</strong> to allocate seats to the group.</li>
	<li><code>boolean scatter(int k, int maxRow)</code> Returns <code>true</code> if all <code>k</code> members of the group can be allocated seats in rows <code>0</code> to <code>maxRow</code>, who may or <strong>may not</strong> sit together. If the seats can be allocated, it allocates <code>k</code> seats to the group with the <strong>smallest</strong> row numbers, and the smallest possible seat numbers in each row. Otherwise, returns <code>false</code>.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Example 1:</strong></p>

<pre>
<strong>Input</strong>
[&quot;BookMyShow&quot;, &quot;gather&quot;, &quot;gather&quot;, &quot;scatter&quot;, &quot;scatter&quot;]
[[2, 5], [4, 0], [2, 0], [5, 1], [5, 1]]
<strong>Output</strong>
[null, [0, 0], [], true, false]

<strong>Explanation</strong>
BookMyShow bms = new BookMyShow(2, 5); // There are 2 rows with 5 seats each 
bms.gather(4, 0); // return [0, 0]
                  // The group books seats [0, 3] of row 0. 
bms.gather(2, 0); // return []
                  // There is only 1 seat left in row 0,
                  // so it is not possible to book 2 consecutive seats. 
bms.scatter(5, 1); // return True
                   // The group books seat 4 of row 0 and seats [0, 3] of row 1. 
bms.scatter(5, 1); // return False
                   // There is only one seat left in the hall.
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= m, k &lt;= 10<sup>9</sup></code></li>
	<li><code>0 &lt;= maxRow &lt;= n - 1</code></li>
	<li>At most <code>5 * 10<sup>4</sup></code> calls <strong>in total</strong> will be made to <code>gather</code> and <code>scatter</code>.</li>
</ul>