<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 primePalindrome(int n) {
        if(n==1)
        return 2;

        while(1)
        {
            if(10000000<=n && n<100000000)
            n=100000000;
            int x=n;
            int sum=0,flag=0;

            while(x)
            {
               sum=(sum*10)+(x%10);
               x/=10; 

            }
            if(sum==n)
            {
                for(int i=2;i<n;i++)
                {
                    if(n%i==0)
                    {
                        flag=1;
                        break;
                    }
                }
            }
            if(sum==n && flag==0)
            break;
            n++;
        }
        return n;
        

        
    }
};
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