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>
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define fi first
#define se second
using namespace std;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> ii;
typedef pair<double,double> dd;
typedef pair<char,char> cc;
typedef vector<ii> vii;
typedef long long ll;
typedef pair<ll, ll> l4;
typedef long double ld;
typedef pair<ld,ld> ldd;
const ld pi = acosl(-1);
const int maxn = 2010;
const ll mod = 998244353;
const ll inf = 1e18;
const ld eps = 1e-10;

string s;
int main() {
    cin >> s;
    int len = s.length();
    int ok = len%2;
    if (len%2 == 0) {
        char c = s[len/2];
        s = s.substr(0,len/2)+c+s.substr(len/2);
    }
    char c = s[len/2];
    int i = 0;
    while (i <= len/2 && s[len/2-i]==c && s[len/2+i]==c) i++;
    i--;
    printf("%d\n",len/2+ok+i);
}