<p>You are given two numeric strings <code>num1</code> and <code>num2</code> and two integers <code>max_sum</code> and <code>min_sum</code>. We denote an integer <code>x</code> to be <em>good</em> if:</p>

<ul>
	<li><code>num1 &lt;= x &lt;= num2</code></li>
	<li><code>min_sum &lt;= digit_sum(x) &lt;= max_sum</code>.</li>
</ul>

<p>Return <em>the number of good integers</em>. Since the answer may be large, return it modulo <code>10<sup>9</sup> + 7</code>.</p>

<p>Note that <code>digit_sum(x)</code> denotes the sum of the digits of <code>x</code>.</p>

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

<pre>
<strong>Input:</strong> num1 = &quot;1&quot;, num2 = &quot;12&quot;, <code>min_sum</code> = 1, max_sum = 8
<strong>Output:</strong> 11
<strong>Explanation:</strong> There are 11 integers whose sum of digits lies between 1 and 8 are 1,2,3,4,5,6,7,8,10,11, and 12. Thus, we return 11.
</pre>

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

<pre>
<strong>Input:</strong> num1 = &quot;1&quot;, num2 = &quot;5&quot;, <code>min_sum</code> = 1, max_sum = 5
<strong>Output:</strong> 5
<strong>Explanation:</strong> The 5 integers whose sum of digits lies between 1 and 5 are 1,2,3,4, and 5. Thus, we return 5.
</pre>

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

<ul>
	<li><code>1 &lt;= num1 &lt;= num2 &lt;= 10<sup>22</sup></code></li>
	<li><code>1 &lt;= min_sum &lt;= max_sum &lt;= 400</code></li>
</ul>

class Solution:
    # digit dp technique
    # let f(idx, bounded, sum) = sum of digits, where bounded reprs whether this index is bounded
    # f(idx, bounded, sum) = f(idx - 1, True, sum + n) if bounded and (n == num2[idx])
    def count(self, num1: str, num2: str, min_sum: int, max_sum: int) -> int:
        num1 = num1.zfill(len(num2)).replace(" ", "0")
        
        @lru_cache(None)
        def f(idx, upper_bounded, lower_bounded, sm):
            if idx >= len(num2): 
                return 1 if min_sum <= sm <= max_sum else 0
            
            upper_bound = ord(num2[idx]) - ord('0') if upper_bounded else 9
            lower_bound = ord(num1[idx]) - ord('0') if lower_bounded else 0
            
            total = 0
            
            for n in range(lower_bound, upper_bound + 1):
                new_upper_bounded = upper_bounded and ord(num2[idx]) - ord('0') == n
                new_lower_bounded = lower_bounded and ord(num1[idx]) - ord('0') == n
                
                if sm + n <= max_sum:
                    total = (total + f(idx + 1, new_upper_bounded, new_lower_bounded, sm + n)) % 1000000007
                
            return total
        
        return f(0, True, True, 0)
                
                
17900