<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 {
    bool isPalindrome(int a){
        int t = a;
        int b = 0;
        while(a>0){
            b = (b*10)+(a%10);
            a/=10;
        }
        return t==b;
    }
    bool isPrime(int a){
        if(a<2)
        return false;
        for(int i=2;i*i<=a;i++){
            if(a%i==0)
            return false;
        }
        return true;
    }
public:
    int primePalindrome(int n) {
        int i=n;
        while(1){
            if(isPalindrome(i) && isPrime(i))
            return i;
            i++;
        }
        return -1;
    }
};
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