<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 {
public:
    int isReverse(int n) {
        int rev = 0;
        while (n > 0) {
            rev = rev * 10 + (n % 10);
            n /= 10;
        }
        return rev;
    }

    bool isPrime(int n) {
        if (n <= 1) return false;
        for (int i = 2; i <= sqrt(n); i++) {
            if (n % i == 0) return false;
        }
        return true;
    }

    bool isPalindrome(int n) {
        return n == isReverse(n);
    }

    int primePalindrome(int n) {
        if (n <= 2) return 2;
        
        for (int i = n; i < 2 * pow(10, 8); i++) {
            if (isPalindrome(i) && isPrime(i)) {
                return i;
            }
            
            if (i > 11 && i == isReverse(i)) {
                int len = log10(i) + 1;
                if (len % 2 == 0) {
                    i = pow(10, len) - 1;
                }
            }
        }
        return -1;
    }
};

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