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 <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <utility>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#define mp make_pair
#define pb push_back
using namespace std; 
typedef long long ll; 
char str[100005]; 
int arr[100005]; 
int main()
{
	// freopen("ARC088-D.in", "r", stdin); 
	scanf("%s", str); 
	int n = strlen(str); 
	for (int i = 0; i < n; i++)
		str[i] -= '0'; 
	for (int i = 0; i <= n; i++)
		arr[i] = str[i] ^ (i ? str[i - 1] : 0); 
	int l = 1, r = n, ans; 
	while (l <= r)
	{
		int m = l + r >> 1, x = 0, y = 0, z = 0; 
		for (int i = m; i <= n; i++)
			x ^= arr[i]; 
		for (int i = 0; i <= min(m - 1, n - m); i++)
			y ^= arr[i]; 
		for (int i = n - m + 1; i < m; i++)
			z += arr[i]; 
		if (x == y && !z)
		{
			ans = m; 
			l = m + 1;
		}
		else
			r = m - 1; 
	}
	printf("%d\n", ans);
	return 0; 
}