<p>You are given positive integers <code>low</code>, <code>high</code>, and <code>k</code>.</p>

<p>A number is <strong>beautiful</strong> if it meets both of the following conditions:</p>

<ul>
	<li>The count of even digits in the number is equal to the count of odd digits.</li>
	<li>The number is divisible by <code>k</code>.</li>
</ul>

<p>Return <em>the number of beautiful integers in the range</em> <code>[low, high]</code>.</p>

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

<pre>
<strong>Input:</strong> low = 10, high = 20, k = 3
<strong>Output:</strong> 2
<strong>Explanation:</strong> There are 2 beautiful integers in the given range: [12,18]. 
- 12 is beautiful because it contains 1 odd digit and 1 even digit, and is divisible by k = 3.
- 18 is beautiful because it contains 1 odd digit and 1 even digit, and is divisible by k = 3.
Additionally we can see that:
- 16 is not beautiful because it is not divisible by k = 3.
- 15 is not beautiful because it does not contain equal counts even and odd digits.
It can be shown that there are only 2 beautiful integers in the given range.
</pre>

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

<pre>
<strong>Input:</strong> low = 1, high = 10, k = 1
<strong>Output:</strong> 1
<strong>Explanation:</strong> There is 1 beautiful integer in the given range: [10].
- 10 is beautiful because it contains 1 odd digit and 1 even digit, and is divisible by k = 1.
It can be shown that there is only 1 beautiful integer in the given range.
</pre>

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

<pre>
<strong>Input:</strong> low = 5, high = 5, k = 2
<strong>Output:</strong> 0
<strong>Explanation:</strong> There are 0 beautiful integers in the given range.
- 5 is not beautiful because it is not divisible by k = 2 and it does not contain equal even and odd digits.
</pre>

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

<ul>
	<li><code>0 &lt; low &lt;= high &lt;= 10<sup>9</sup></code></li>
	<li><code>0 &lt; k &lt;= 20</code></li>
</ul>

class Solution:
    def numberOfBeautifulIntegers(self, low: int, high: int, k: int) -> int:
        
        @cache
        def dfs(s, index, odd, even, remainder, tight, leadingZero):
            if index >= len(s):
                return remainder % k == 0 and odd == even
            
            bound = int(s[index]) if tight else 9
            ans = 0
            for digit in range(bound + 1):
                add_odd = digit % 2 == 1
                add_even = digit % 2 == 0
                
                if leadingZero and digit == 0:
                    add_even = 0
                    
                ans += dfs(s, index + 1,
                           odd + add_odd,
                           even + add_even,
                           (remainder*10 + digit) % k,
                           tight and digit == int(s[index]),
                           leadingZero and digit == 0)
            return ans
        
        return dfs(str(high), 0, 0, 0, 0, True, True) - dfs(str(low - 1), 0, 0, 0, 0, True, True)
21000