<p>Given an integer array <code>queries</code> and a <strong>positive</strong> integer <code>intLength</code>, return <em>an array</em> <code>answer</code> <em>where</em> <code>answer[i]</code> <em>is either the </em><code>queries[i]<sup>th</sup></code> <em>smallest <strong>positive palindrome</strong> of length</em> <code>intLength</code> <em>or</em> <code>-1</code><em> if no such palindrome exists</em>.</p>

<p>A <strong>palindrome</strong> is a number that reads the same backwards and forwards. Palindromes cannot have leading zeros.</p>

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

<pre>
<strong>Input:</strong> queries = [1,2,3,4,5,90], intLength = 3
<strong>Output:</strong> [101,111,121,131,141,999]
<strong>Explanation:</strong>
The first few palindromes of length 3 are:
101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, ...
The 90<sup>th</sup> palindrome of length 3 is 999.
</pre>

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

<pre>
<strong>Input:</strong> queries = [2,4,6], intLength = 4
<strong>Output:</strong> [1111,1331,1551]
<strong>Explanation:</strong>
The first six palindromes of length 4 are:
1001, 1111, 1221, 1331, 1441, and 1551.
</pre>

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

<ul>
	<li><code>1 &lt;= queries.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= queries[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= intLength&nbsp;&lt;= 15</code></li>
</ul>

class Solution:
    def kthPalindrome(self, queries: List[int], intLength: int) -> List[int]:

        digits_In_First_Half = (intLength // 2) + (intLength % 2)
        # 6 -> 3 + 0; 5 -> 2 + 1 = 3.
        first_Number_With_D_digits = 10 ** (digits_In_First_Half - 1)   # 10^(2-1) = 10
        last_Number_With_D_digits = (10 ** digits_In_First_Half) - 1    # 10*2 - 1 = 99
        
        def make_palindrome(N: int, L: int) -> int:
            digits = list(str(N))
            rDigits = list(reversed(digits))
            if (len(digits) * 2) - 1 == L:
                rDigits = rDigits[1:]       # throw out the first digit
            answer = digits + rDigits
            if len(answer) != L:
                print(f'ERROR: {len(answer)=} != {L=}\n\t{answer=}')
                return -99
            return int(''.join(answer))

        def doQuery(N: int) -> int:
            Nth_number = first_Number_With_D_digits + N - 1
            if Nth_number > last_Number_With_D_digits:
                return -1

            return make_palindrome(Nth_number, intLength)
        
        return [
            doQuery(Q)
            for Q in queries
        ]

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