human
I found an interesting problem on https://leetcode.com/problems/prime-palindrome/:
Find the smallest prime palindrome greater than or equal to N.
Recall that a number is prime if it's only divisors are 1 and itself, and it is greater than 1. 
For example, 2,3,5,7,11 and 13 are primes.
Recall that a number is a palindrome if it reads the same from left to right as it does from right to left. 
For example, 12321 is a palindrome.
 

Example 1:
Input: 6
Output: 7

Example 2:
Input: 8
Output: 11

Example 3:
Input: 13
Output: 101


 
Note:

1 <= N <= 10^8
The answer is guaranteed to exist and be less than 2 * 10^8.

I tried it in Python, but could not do it. Can you solve it?
assistant
#Here is Python code to find the smallest prime palindrome greater than or equal to N.

class Solution:
    def primePalindrome(self, N: int) -> int:
        def isPrime(N):
            return N > 1 and all(N % d for d in range(2, int(N**0.5)+1))
        
        # N must be a palindrome with odd number of digits.
        # The return value will have odd number of digits too.
        def nextPalindrome(N):
            if N in [999, 99999, 9999999]:
                return (N + 1) * 10 + 1
            n = str(N // 10 ** (len(str(N))//2) + 1)
            return int(n + n[-2::-1])
        
        if N <= 11: 
            while not isPrime(N):
                N += 1
            return N
        
        if (digits := len(str(N))) % 2 == 0:
            N = 10 ** digits + 1
        else:
            n = str(N // 10 ** (len(str(N))//2))
            if (p := int(n + n[-2::-1])) >= N:
                N = p
            else:
                N = nextPalindrome(p)
                
        while not isPrime(N):
            N = nextPalindrome(N)
        return N