<p>Let <code>f(x)</code> be the number of zeroes at the end of <code>x!</code>. Recall that <code>x! = 1 * 2 * 3 * ... * x</code> and by convention, <code>0! = 1</code>.</p>

<ul>
	<li>For example, <code>f(3) = 0</code> because <code>3! = 6</code> has no zeroes at the end, while <code>f(11) = 2</code> because <code>11! = 39916800</code> has two zeroes at the end.</li>
</ul>

<p>Given an integer <code>k</code>, return the number of non-negative integers <code>x</code> have the property that <code>f(x) = k</code>.</p>

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

<pre>
<strong>Input:</strong> k = 0
<strong>Output:</strong> 5
<strong>Explanation:</strong> 0!, 1!, 2!, 3!, and 4! end with k = 0 zeroes.
</pre>

<p><strong>Example 2:</strong></p>

<pre>
<strong>Input:</strong> k = 5
<strong>Output:</strong> 0
<strong>Explanation:</strong> There is no x such that x! ends in k = 5 zeroes.
</pre>

<p><strong>Example 3:</strong></p>

<pre>
<strong>Input:</strong> k = 3
<strong>Output:</strong> 5
</pre>

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

<ul>
	<li><code>0 &lt;= k &lt;= 10<sup>9</sup></code></li>
</ul>