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: Wrong Answer)

#include<iostream>
#include<cstring>
using namespace std;
int main(){
	string s;
	cin>>s;
	int l=s.size();
	int ans;
    ans=l/2;
    string s1,s2;
	if (l%2==0){
		s1=s.substr(0,l/2);
		s2=s.substr(l/2,l/2);
		
	//	cout<<s1<<endl<<s2<<endl;
		
		for (int i=0;i<l/2;i++){
			if (s2[i]==s1[l/2-i-1]) {
				ans++;
			}
			else{
				break;
			}
		}
	
		cout<<ans<<endl;
	}else{
		s1=s.substr(0,l/2);
		s2=s.substr(l/2,l/2);
		//cout<<s1<<endl<<s2<<endl;
		
		for (int i=0;i<l/2;i++){
			if (s2[i]==s1[l/2-i-1]) {
				ans++;
			}
			else{
				break;
			}
		}
		
		int ans2=l/2;
		s1=s.substr(1,l/2);
		s2=s.substr(l/2+1,l/2);
		//cout<<s1<<endl<<s2<<endl;
		
		for (int i=0;i<l/2;i++){
			if (s2[i]==s1[l/2-i-1]) {
				ans2++;
			}
			else{
				break;
			}
		}
		if (ans2>ans)ans=ans2;
		
		
		ans2=l/2;
		s1=s.substr(0,l/2);
		s2=s.substr(l/2+1,l/2);
		//cout<<s1<<endl<<s2<<endl;
		
		for (int i=0;i<l/2;i++){
			if (s2[i]==s1[l/2-i-1]) {
				ans2++;
			}
			else{
				break;
			}
		}
		if (ans2>ans)ans=ans2;
		
		cout<<ans<<endl;
		
	} 

}


/*
l        k
1        1
2        1

3              111   3
               110   2
			   101   2
			   
			   
100000000
			   
			   
k=l        改不了 
k=l-1      改得了头尾  
k=l-2      改得了头尾，改得了 第二倒二
k=l-3       



*/