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;

#define fs first
#define sc second
#define pb emplace_back
#define mp make_pair
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()

const int32_t inf = 1001001001;
const int64_t infll = 1001001001001001001ll;
const int dx[] = {0, -1, 1, 0, -1, 1, -1, 1}, dy[] = {-1, 0, 0, 1, -1, -1, 1, 1};

template <typename T> using vector2d = vector<vector<T>>;

// clang-format off
template <typename T> void sort(vector<T> &v) { sort(all(v)); }
ostream &operator<<(ostream &os, __int128_t value) { if (ostream::sentry(os)) { __uint128_t tmp = value < 0 ? -value : value; char buffer[64]; char *d = end(buffer); do { --d; *d = "0123456789"[tmp % 10]; tmp /= 10; } while (tmp != 0); if (value < 0) { --d; *d = '-'; } int len = end(buffer) - d; if (os.rdbuf()->sputn(d, len) != len) { os.setstate(ios_base::badbit); }} return os; }
istream &operator>>(istream &is, __int128_t &value) { string in; is >> in; value = 0; for (const char &c : in) { if ('0' <= c && c <= '9') value = 10 * value + (c - '0'); } if (in[0] == '-') value *= -1; return is; }
ostream &operator<<(ostream &os, __uint128_t value) { if (ostream::sentry(os)) { char buffer[64]; char *d = end(buffer); do { --d; *d = "0123456789"[value % 10]; value /= 10; } while (value != 0); int len = end(buffer) - d; if (os.rdbuf()->sputn(d, len) != len) { os.setstate(ios_base::badbit); }} return os; }
istream &operator>>(istream &is, __uint128_t &value) { string in; is >> in; value = 0; for (const char &c : in) { if ('0' <= c && c <= '9') value = 10 * value + (c - '0'); } return is; }
template <typename T> ostream &operator<<(ostream &os, vector<T> &v) { os << v[0]; for (int i = 1; i < v.size(); ++i) os << " " << v[i]; return os; }
template <typename T> istream &operator>>(istream &is, vector<T> &v) { for (T &in : v) is >> in; return is; }
template <typename T1, typename T2> ostream &operator<<(ostream &os, pair<T1, T2> p) { os << p.fs << " " << p.sc; return os; }
template <typename T1, typename T2> istream &operator>>(istream &is, pair<T1, T2> &p) { is >> p.fs >> p.sc; return is; }

struct IoSetup { IoSetup(){ cin.tie(0); ios::sync_with_stdio(0); cout << fixed << setprecision(10); cerr << fixed << setprecision(10); } } iosetup;

inline int64_t in() { int64_t x = 0; cin >> x; return x; }
template <typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { return a < b && (a = b, true); }
template <typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { return a > b && (a = b, true); }
// clang-format on

signed main(int argc, char *argv[]) {
  string s;
  cin >> s;


  vector<int> a;
  int cnt = 1;
  for (int i = 1; i < s.size(); ++i) {
    if (s[i] != s[i - 1]) {
      a.pb(cnt);
      cnt = 0;
    }
    cnt++;
  }
  a.pb(cnt);


  if (a.size() == 1) {
    cout << a[0] << endl;
  } else if (a.size() == 2) {
    cout << max(a[0], a[1]) << endl;
  } else {
    int l = s[0] == '0';
    int r = a.size() - (s.back() == '0');
    int ans = inf;
    for (int i = l; i <= r - 3; ++i) {
      chmin(ans, a[i] + a[i + 1] + a[i + 2]);
    }
    int t = 0;
    for (int i = 0; i < a.size() - 1; ++i) {
      chmax(t, a[i] + a[i + 1]);
    }
    cout << min(ans, t) << endl;
  }
}