<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 {
public:
    vector<long long> kthPalindrome(vector<int>& queries, int intLength) {
        int n = queries.size();
        int m = (intLength + 1) / 2;
        long long basic = 1;
        for (int i = 1; i < m; ++i) {
            basic *= 10;
        }
        vector<long long> ans;
        for (int i = 0; i < n; ++i) {
            int query = queries[i];
            long long res = basic + query - 1; 
            if (res >= basic * 10) {
                ans.push_back(-1);
                continue;
            }
            int digits[15];
            int k = m - 1;
            while (res > 0) {
                digits[k] = res % 10;
                res /= 10;
                k--;
            }
            for (int j = m; j < intLength; ++j) {
                digits[j] = digits[intLength - 1 - j];
            }
            res = 0;
            for (int j = 0; j < intLength; ++j) {
                res = res * 10 + digits[j];
            }
            ans.push_back(res);
        }
        return ans;
    }
    // intLength - 1, ..., 0
};
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