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<bits/stdc++.h>
using namespace std;
typedef long long ll;
template<typename T1,typename T2> bool chmin(T1 &a,T2 b){if(a<=b)return 0; a=b; return 1;}
template<typename T1,typename T2> bool chmax(T1 &a,T2 b){if(a>=b)return 0; a=b; return 1;}


signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << fixed << setprecision(20);

    
    string s;
    cin>>s;
    int ans = s.size();
    int n = s.size();
    int cnt0=0, cnt1=0;
    for(int i=0;i<n;i++){
        if(s[i] == '0'){
            if(i && s[i-1]=='1'){
                if(cnt0) chmin(ans,cnt1+cnt0);
                else chmin(ans,max(i,n-i));
                cnt0 = 0;
            }
            cnt0++;
        }
        else{
            if(i && s[i-1]=='0'){
                if(cnt1) chmin(ans,cnt1+cnt0);
                else chmin(ans,max(i,n-i));
                cnt1=0;
            }
            cnt1++;
        }
    }
    if(cnt0 && cnt1) chmin(ans,cnt0+cnt1);
    else ans = n;
    cout << ans << endl;

}