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++ (GCC 9.2.1) (judge verdict: Wrong Answer)


// Problem : D - Wide Flip
// Contest : AtCoder - AtCoder Regular Contest 088
// URL : https://atcoder.jp/contests/arc088/tasks/arc088_b
// Memory Limit : 256 MB
// Time Limit : 2000 ms
// Powered by CP Editor (https://github.com/cpeditor/cpeditor)

/*⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀
⠀⠀⠀⠀⣠⣶⡾⠏⠉⠙⠳⢦⡀⠀⠀⠀
⠀⠀⠀⣴⠿⠏⠀⠀⠀⠀⠀⠀⢳⡀⠀
⠀⠀⢠⣟⣋⡀⢀⣀⣀⡀⠀⣀⡀⣧⠀
⠀⠀⢸⣯⡭⠁⠸⣛⣟⠆⡴⣻⡲⣿⠀
⠀⠀⣟⣿⡭⠀⠀⠀⠀⠀⢱⠀⠀⣿⠀
⠀⠀⠙⢿⣯⠄⠀⠀⠀⢀⡀⠀⠀⡿⠀⠀
⠀⠀⠀⠀⠹⣶⠆⠀⠀⠀⠀⠀⡴⠃⠀⠀
⠀⠀⠀⠀⠀⢸⣷⡦⢤⡤⢤⣞⣁⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀
⠀⠀⢀⣤⣴⣿⣏⠁⠀⠀⠸⣏⢯⣷⣖⣦⡀⠀⠀⠀⠀⠀⠀
⢀⣾⣽⣿⣿⣿⣿⠛⢲⣶⣾⢉⡷⣿⣿⠵⣿⠀⠀⠀⠀⠀⠀
⣼⣿⠍⠉⣿⡭⠉⠙⢺⣇⣼⡏⠀⠀⠀⣄⢸⠀⠀⠀⠀⠀⠀
⣿⣿⣧⣀⣿.........⣀⣰⣏⣘⣆⣀⠀
*/
#include <bits/stdc++.h> 
#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
 
#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
 
 
using namespace std;
using namespace __gnu_pbds;
 
typedef long long   ll;
typedef long double ld;
typedef complex< ld > cd;
 
typedef pair<int, int> pi;
typedef pair<ll, ll>   pl;
typedef pair<ld, ld>   pd;
 
typedef vector< int > vi;
typedef vector< ld >  vd;
typedef vector< ll >  vl;
typedef vector< pi >  vpi;
typedef vector< pl >  vpl;
typedef vector< cd >  vcd;
 
#define l00p(i, a, b) for (int i=a; i<(b); i++)
#define loop(i, a)    for (int i=0; i<(a); i++)
#define rep1(i, a, b) for (int i = (b)-1; i >= a; i--)
#define rep(i, a)     for (int i = (a)-1; i >= 0; i--)
#define trav(a, x)    for (auto& a : x)
#define uid(a, b)     uniform_int_distribution<int>(a, b)(rng)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f  first
#define s  second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
#define ins insert
//#define MOD 998244353
 
template<class T> bool ckmin(T& a, const T& b) { return b < a ? a = b, 1 : 0; }
template<class T> bool ckmax(T& a, const T& b) { return a < b ? a = b, 1 : 0; }

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const ld pii=3.14159265359; 
const int MOD = 1000000007;
const char nl = '\n';
const ll INF = 1e18;
const int MX = 100001; //check the limits, dummy
string s;
bool check(int k)
{
	bool z=true,o=true;
	loop(i,sz(s))
	{
		if(i+k<sz(s)||i-k>=0)
		continue;
		z&=s[i]=='0';
		o&=s[i]=='1';
	}
	return z||o;
}
int main() {
	ios_base::sync_with_stdio(0); cin.tie(0); 
	cin>>s;
	int n=sz(s);
	int l=1,r=n;
	int ans=-1;
	while(l<r)
	{
		int mid=(l+r)/2;
		if(check(mid))
		{
			ckmax(ans,mid);
			l=mid+1;
			}
		else
		r=mid-1;
	}
	cout<<ans<<nl;
	return 0;
}
// read the question correctly (ll vs int)
// template by bqi343