<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]:
        def generate_palindrome(k):
            if intLength==1:
                return k if 0<k<10 else -1
            if intLength%2==0:
                size = intLength//2
                start = '1'+('0'*(size-1))
                end   = '9'+('9'*(size-1))
                available = int(end)-int(start)+1
                if k>available: return -1
                left = str(int(start) + (k-1))
                return int(left+left[::-1])
            else:  # intLength%2==1
                size = (intLength+1)//2
                start = '1'+('0'*(size-1))
                end   = '9'+('9'*(size-1))
                available = int(end)-int(start)+1
                if k>available: return -1
                left = str(int(start) + (k-1))
                return int(left+left[:-1][::-1])
        return [generate_palindrome(k) for k in queries]

        
498