<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:
    bool ispalindrome(int n){
        string s = to_string(n);
        int l = 0, r = s.size()-1;
        while(l<r){
            if(s[l] != s[r]){
                return false;
            }
            l++;r--;
        }
        return true;
    }
    bool isp(int n){
        if (n <= 1){
            return false;
        }
        if(n == 2 || n== 3 || n == 5){
            return true;
        }

        for(int i = 2; i*i <= n; i++){
            if(n%i == 0){
                return false;
            }
        }

        return true;
    }


    int primePalindrome(int n) {
        int k = n==1?2:n;
        while(true){
            if((k > 1e3 && k< 1e4) || (k>1e5 && k < 1e6) || k> 1e7 && k<1e8){
                k = pow(10, ceil(log10(k)));
                continue;
            }
            if(isp(k) && ispalindrome(k)){
                return k;
            }
            k++;
        }
        return -1;
    }
};
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