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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

//~ #include <cmath>
//~ #include <cassert>
#include <iostream>
//~ #include <iomanip>
//~ #include <map>
//~ #include <set>
//~ #include <vector>
//~ #include <algorithm>
#include <queue>
//~ #include <cstring>

using namespace std;

#define forsn(i,s,n) for (int i = (s);i < (int)(n);i++)
#define forn(i,n) forsn(i,0,n)
#define dforsn(i,s,n) for (int i = (n-1);i >= (int)(s);i--)
#define dforn(i,n) dforsn(i,0,n)
#define all(v) (v).begin(), (v).end()
#define F first
#define S second
#define dbg(x) cerr << #x << " = " << x << endl

string s; 

//100000000

bool doable(int size) {
	int from_left = 2 * size;
	deque<int> current;
	
	forn(i, s.size()) {
		if (!current.empty() && (i - current.front()) == size) { current.pop_front(); from_left++; }
		if (((s[i] == '1') ^ (((int)current.size()) % 2)) ^ (from_left % 2)) {
			if (i + size <= (int)s.size()) {
				current.push_back(i);
			} else {
				if (from_left && i >= size) {
					from_left--;
				} else {
					return false;
				}
			}
		}
	}
	
	return true;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(NULL);

	cin >> s;
	
	int left = 1, right = s.size() + 1;
	while (left + 1 < right) {
		int middle = (left + right) / 2;
		if (doable(middle)) {
			left = middle;
		} else {
			right = middle;
		}
	}
	
	cout << left << endl;
}