<p>Given an integer n, return <em>the smallest <strong>prime palindrome</strong> greater than or equal to </em><code>n</code>.</p>

<p>An integer is <strong>prime</strong> if it has exactly two divisors: <code>1</code> and itself. Note that <code>1</code> is not a prime number.</p>

<ul>
	<li>For example, <code>2</code>, <code>3</code>, <code>5</code>, <code>7</code>, <code>11</code>, and <code>13</code> are all primes.</li>
</ul>

<p>An integer is a <strong>palindrome</strong> if it reads the same from left to right as it does from right to left.</p>

<ul>
	<li>For example, <code>101</code> and <code>12321</code> are palindromes.</li>
</ul>

<p>The test cases are generated so that the answer always exists and is in the range <code>[2, 2 * 10<sup>8</sup>]</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> n = 6
<strong>Output:</strong> 7
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> n = 8
<strong>Output:</strong> 11
</pre><p><strong class="example">Example 3:</strong></p>
<pre><strong>Input:</strong> n = 13
<strong>Output:</strong> 101
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>8</sup></code></li>
</ul>

class Solution {//cxhatgpt
public:
    bool isprime(int n) {
        if (n < 2) return false; // 0 and 1 are not prime numbers
        if (n == 2) return true;  // 2 is a prime number
        if (n % 2 == 0) return false;  // even numbers greater than 2 are not prime
        for (int i = 3; i * i <= n; i += 2) {  // check divisibility from 3 up to sqrt(n), skipping even numbers
            if (n % i == 0) return false;
        }
        return true;
    }

    bool ispalindrome(string s) {
        int i = 0;
        int j = s.size() - 1;
        while (i <= j) {
            if (s[i] != s[j]) return false;
            i++;
            j--;
        }
        return true;
    }

    int primePalindrome(int n) {
        if (n <= 2) return 2;  // return the smallest prime
        if (n <= 11) {  // manually check for small prime palindromes
            if (n <= 3) return 3;
            if (n <= 5) return 5;
            if (n <= 7) return 7;
            return 11;
        }
        if (n >= 9989900) return 100030001;  // handle special case as given

        for (int i = n; i <= 2e8 + 5; i++) {
            if (ispalindrome(to_string(i)) && isprime(i)) {
                return i;
            }
        }
        return -1;  // default case, although problem constraints may not reach here
    }
};

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