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 Rust (1.15.1) (judge verdict: Accepted)

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}

fn main() {
    let s: String = read();
    let s = s.chars().collect::<Vec<_>>();
    let central_char = s[s.len() / 2];

    let mut former_idx;
    let mut latter_idx;
    if s.len() % 2 == 0 {
        former_idx = s.len() / 2 - 1;
        latter_idx = s.len() / 2;
    } else {
        former_idx = s.len() / 2;
        latter_idx = s.len() / 2;
    }

    let mut ans = s.len() / 2;
    loop {
        if s[former_idx] != central_char || s[latter_idx] != central_char {
            break;
        }
        ans += 1;
        if former_idx == 0 {
            break;
        }
        latter_idx += 1;
        former_idx -= 1;
    }
    println!("{}", ans);
}