Problem:
Score : 500 points

Problem Statement

You are given a string S consisting of 0 and 1.
Find the maximum integer K not greater than |S| such that we can turn all the characters of S into 0 by repeating the following operation some number of times.

Choose a contiguous segment [l,r] in S whose length is at least K (that is, r-l+1\geq K must be satisfied). For each integer i such that l\leq i\leq r, do the following: if S_i is 0, replace it with 1; if S_i is 1, replace it with 0.

Constraints

1\leq |S|\leq 10^5

S_i(1\leq i\leq N) is either 0 or 1.

Input

Input is given from Standard Input in the following format:

S

Output

Print the maximum integer K such that we can turn all the characters of S into 0 by repeating the operation some number of times.

Sample Input 1

010

Sample Output 1

2

We can turn all the characters of S into 0 by the following operations:

Perform the operation on the segment S[1,3] with length 3. S is now 101.

Perform the operation on the segment S[1,2] with length 2. S is now 011.

Perform the operation on the segment S[2,3] with length 2. S is now 000.

Sample Input 2

100000000

Sample Output 2

8

Sample Input 3

00001111

Sample Output 3

4

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;
import java.io.*;
import java.nio.charset.StandardCharsets;

public class Main {
	public static void main(String[] args) throws IOException {
		InputStreamReader reader = new InputStreamReader(System.in, StandardCharsets.UTF_8);
		BufferedReader in = new BufferedReader(reader);

		String S = in.readLine();
		Deque<Integer> queue = new ArrayDeque<>();
		queue.addLast(1);
		for (int i = 1; i < S.length(); ++i) {
			if (S.charAt(i - 1) != S.charAt(i)) {
				queue.addLast(1);
			} else {
				queue.addLast(queue.pollLast() + 1);
			}
		}
		List<Integer> list = new ArrayList<>();
		list.addAll(queue);

		int minLen = Integer.MAX_VALUE;
		int maxLen = Integer.MIN_VALUE;
		for (Integer i : list) {
			minLen = Math.min(minLen, i);
			maxLen = Math.max(maxLen, i);
		}
		if (list.size() <= 2) {
			// maxLen
			System.out.println(maxLen);
		} else {
			System.out.println(minLen * 2);
		}
	}
}