<p>You are given two positive integers <code>left</code> and <code>right</code> with <code>left &lt;= right</code>. Calculate the <strong>product</strong> of all integers in the <strong>inclusive</strong> range <code>[left, right]</code>.</p>

<p>Since the product may be very large, you will <strong>abbreviate</strong> it following these steps:</p>

<ol>
	<li>Count all <strong>trailing</strong> zeros in the product and <strong>remove</strong> them. Let us denote this count as <code>C</code>.

	<ul>
		<li>For example, there are <code>3</code> trailing zeros in <code>1000</code>, and there are <code>0</code> trailing zeros in <code>546</code>.</li>
	</ul>
	</li>
	<li>Denote the remaining number of digits in the product as <code>d</code>. If <code>d &gt; 10</code>, then express the product as <code>&lt;pre&gt;...&lt;suf&gt;</code> where <code>&lt;pre&gt;</code> denotes the <strong>first</strong> <code>5</code> digits of the product, and <code>&lt;suf&gt;</code> denotes the <strong>last</strong> <code>5</code> digits of the product <strong>after</strong> removing all trailing zeros. If <code>d &lt;= 10</code>, we keep it unchanged.
	<ul>
		<li>For example, we express <code>1234567654321</code> as <code>12345...54321</code>, but <code>1234567</code> is represented as <code>1234567</code>.</li>
	</ul>
	</li>
	<li>Finally, represent the product as a <strong>string</strong> <code>&quot;&lt;pre&gt;...&lt;suf&gt;eC&quot;</code>.
	<ul>
		<li>For example, <code>12345678987600000</code> will be represented as <code>&quot;12345...89876e5&quot;</code>.</li>
	</ul>
	</li>
</ol>

<p>Return <em>a string denoting the <strong>abbreviated product</strong> of all integers in the <strong>inclusive</strong> range</em> <code>[left, right]</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> left = 1, right = 4
<strong>Output:</strong> &quot;24e0&quot;
<strong>Explanation:</strong> The product is 1 &times; 2 &times; 3 &times; 4 = 24.
There are no trailing zeros, so 24 remains the same. The abbreviation will end with &quot;e0&quot;.
Since the number of digits is 2, which is less than 10, we do not have to abbreviate it further.
Thus, the final representation is &quot;24e0&quot;.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> left = 2, right = 11
<strong>Output:</strong> &quot;399168e2&quot;
<strong>Explanation:</strong> The product is 39916800.
There are 2 trailing zeros, which we remove to get 399168. The abbreviation will end with &quot;e2&quot;.
The number of digits after removing the trailing zeros is 6, so we do not abbreviate it further.
Hence, the abbreviated product is &quot;399168e2&quot;.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> left = 371, right = 375
<strong>Output:</strong> &quot;7219856259e3&quot;
<strong>Explanation:</strong> The product is 7219856259000.
</pre>

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

<ul>
	<li><code>1 &lt;= left &lt;= right &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution:
    def abbreviateProduct(self, left: int, right: int) -> str:
        product, suffix, zeros, org_digits = 1.0, 1, 0, 0  
        for n in range(left, right + 1):  
            product *= n  
            while product >= 1:  
                product /= 10  
                org_digits += 1  
            suffix *= n  
            while suffix % 10 == 0:  
                zeros += 1  
                suffix //= 10  
            if suffix > 10 ** 14:  
                suffix %= 10 ** 14  
        if org_digits - zeros <= 10:  
            return str(int(product * (10 ** (org_digits - zeros)) + 0.5)) + 'e' + str(zeros)  
        else:  
            return str(int(product * 100000)) + '...' + ('0000' + str(suffix))[-5:] + 'e' + str(zeros)  

    def factorial(self, n):  
        """Returns the factorial of n."""  
        if n < 0:  
            raise ValueError("Factorial is not defined for negative numbers")  
        if n == 0 or n == 1:  
            return 1  
        result = 1  
        for i in range(2, n + 1):  
            result *= i  
        return result  

    def is_product_large(self, left, right, threshold=10**14):  
        """Checks if the product of the range left to right exceeds the given threshold."""  
        product = 1  
        for n in range(left, right + 1):  
            product *= n  
            if product > threshold:  
                return True  
        return False        
517