<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]:
        palLen = intLength // 2 + (intLength & 1)
        minNum = 10 ** (palLen - 1)
        print(palLen, minNum)

        def find(query):
            num = minNum + query - 1
            numStr = str(num)
            res = numStr + numStr[::-1]
            if intLength & 1:
                res = numStr[:-1] + numStr[::-1]

            if len(res) > intLength:
                return -1
            else:
                return int(res)

        return map(find, queries)

24000