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	`1`	`+class Solution:`
	`2`	`+ def nearestPalindromic(self, n: str) -> str:`
	`3`	`+ length = len(n)`
	`4`	`+ num = int(n)`
	`5`	`+`
	`6`	`+ # Edge cases for very small numbers`
	`7`	`+ if num == 0:`
	`8`	`+ return "1"`
	`9`	`+ if num == 1:`
	`10`	`+ return "0"`
	`11`	`+`
	`12`	`+ # Handle corner cases like "1000", "999"`
	`13`	`+ candidates = set()`
	`14`	`+ candidates.add(str(10**(length - 1) - 1)) # largest number with one less digit (e.g., 999 for 1000)`
	`15`	`+ candidates.add(str(10**length + 1)) # smallest number with one more digit (e.g., 10001 for 9999)`
	`16`	`+`
	`17`	`+ # Generate the palindrome by mirroring`
	`18`	`+ prefix = int(n[:(length + 1) // 2])`
	`19`	`+ for i in [-1, 0, 1]:`
	`20`	`+ new_prefix = str(prefix + i)`
	`21`	`+ if length % 2 == 0:`
	`22`	`+ candidate = new_prefix + new_prefix[::-1]`
	`23`	`+ else:`
	`24`	`+ candidate = new_prefix + new_prefix[:-1][::-1]`
	`25`	`+ candidates.add(candidate)`
	`26`	`+`
	`27`	`+ # Remove the original number from the candidates`
	`28`	`+ candidates.discard(n)`
	`29`	`+`
	`30`	`+ # Find the closest palindrome`
	`31`	`+ closest = min(candidates, key=lambda x: (abs(int(x) - num), int(x)))`
	`32`	`+`
	`33`	`+ return closest`

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