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 <bits/stdc++.h>

using namespace std;

string s;
int ss[100050];
inline int query(int x, int y){ if(x == 0) return ss[y]; else return ss[y]-ss[x-1]; }
int main(){
	cin >> s;
	int n = s.length();
	for(int i = 0; i<n; i++) ss[i] = s[i]-'0';
	for(int i = 1; i<n; i++) ss[i] += ss[i-1];
	if(query(0,n-1) == 0){ cout << n; return 0; }
	int ans = INT_MAX;
	for(int i = 0; i<n; i++){
		if(query(i,i) == 1){
			int cur = 0;
			if(query(0,i) == i+1) cur = max(i+1, cur);
			if(query(i,n-1) == n-i) cur = max(n-i, cur);
			cur = max(i, cur);
			cur = max(n-i-1, cur);
			ans = min(ans, cur);
		}
	}
	for(int i = 0; i<n; i++) ss[i] = s[i]-'0';
	for(int i = 0; i<n; i++) ss[i] = 1-ss[i];
	for(int i = 1; i<n; i++) ss[i] += ss[i-1];
	if(query(0,n-1) == 0){ cout << n; return 0; }
	int ans2 = INT_MAX;
	for(int i = 0; i<n; i++){
		if(query(i,i) == 1){
			int cur = 0;
			if(query(0,i) == i+1) cur = max(i+1, cur);
			if(query(i,n-1) == n-i) cur = max(n-i, cur);
			cur = max(i, cur);
			cur = max(n-i-1, cur);
			ans2 = min(ans2, cur);
		}
	}
	cout << max(ans,ans2);
}