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)

// #define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define rep(i, n) for (int i = 0; i < n; ++i)
#define REP(i, n) for (int i = 0; i < n; ++i)
#define range(i,a,b) ((a)<=(i) && (i)<(b))
#define debug(x) cout << #x << ' ' << '=' << ' ' << (x) << endl;
#define fs first
#define sc second
#define pb push_back
#define eb emplace_back
typedef long long ll;
typedef pair<ll, ll> P;
typedef tuple<ll, ll, ll> T;
typedef vector<ll> vec;
typedef vector<P> pvec;
typedef vector<vector<ll>> vvec;
typedef vector<vector<P>> pvvec;
typedef priority_queue<ll> PQI;
typedef priority_queue<P> PQP;
typedef priority_queue<ll,vector<ll>,greater<ll>> PQIG;
typedef priority_queue<P,vector<P>,greater<P>> PQPG;
const vector<int> dx = {0, -1, 0, 1, 1, 1, -1, -1};
const vector<int> dy = {1, 0, -1, 0, 1, -1, 1, -1};
constexpr int MOD = (1000000007);
// const int MOD = (998244353);
// const int INF = (1 << 30); // 1073741824
const int INF = (1LL << 60); // 1152921504606846976
const double EPS = (1 >> 30);
template<class T> inline bool chmin(T& a, T b) {if (a > b) {a = b; return 1;} return 0;}
template<class T> inline bool chmax(T& a, T b) {if (a < b) {a = b; return 1;} return 0;}
template< typename T1, typename T2 > istream &operator>>(istream &is, pair< T1, T2 > &p) { is >> p.first >> p.second; return is; }
int nmax=200000;
vvec g(nmax);



signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << fixed << setprecision(15);
    //---------------------------------------------
    
    string s;
    cin>>s;
    int n=s.size();
    vec a;
    int cnt=1;
    if(n==1){
        cout<<1<<endl;
        return 0;
    }
    rep(i,n-1){
        if(s[i]==s[i+1]){
            cnt++;
        }
        else{
            a.pb(cnt);
            cnt=1;
        }
        if(i==n-2) a.pb(cnt);
    }
    // rep(i,n) cout<<a[i]<<endl;
    if(a.size()==1){
        cout<<n<<endl;
    }
    else if(a.size()==2){
        cout<<max(a[0],a[1])<<endl;
    }
    else{
        int ans=0;
        rep(i,a.size()-1){
            chmax(ans,a[i]+a[i+1]);
        }
        cout<<ans<<endl;
    }
}