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 int long long
#define pii pair<int,int>
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(c) (c).begin(),(c).end()
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
#define MINF(a) memset(a,0x3f,sizeof(a))
#define POW(n) (1LL<<(n))
#define IN(i,a,b) (a <= i && i <= b)
using namespace std;
template <typename T> inline bool CHMIN(T& a,T b) { if(a>b) { a=b; return 1; } return 0; }
template <typename T> inline bool CHMAX(T& a,T b) { if(a<b) { a=b; return 1; } return 0; }
template <typename T> inline void SORT(T& a) { sort(ALL(a)); }
template <typename T> inline void REV(T& a) { reverse(ALL(a)); }
template <typename T> inline void UNI(T& a) { sort(ALL(a)); a.erase(unique(ALL(a)),a.end()); }
template <typename T> inline T LB(vector<T>& v, T a) { return *lower_bound(ALL(v),a); }
template <typename T> inline int LBP(vector<T>& v, T a) { return lower_bound(ALL(v),a) - v.begin(); }
template <typename T> inline T UB(vector<T>& v, T a) { return *upper_bound(ALL(v),a); }
template <typename T> inline int UBP(vector<T>& v, T a) { return upper_bound(ALL(v),a) - v.begin(); }
template <typename T1, typename T2> ostream& operator<< (ostream& os, const pair<T1,T2>& p) { os << p.first << " " << p.second; return os; }
template <typename T1, typename T2> istream& operator>> (istream& is, pair<T1,T2>& p) { is >> p.first >> p.second; return is; }
template <typename T> ostream& operator<< (ostream& os, const vector<T>& v) { REP(i,v.size()) { if (i) os << " "; 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 T = int> vector<T> make_v(size_t a) { return vector<T>(a); }
template <typename T, typename... Ts> auto make_v(size_t a, Ts... ts) { return vector<decltype(make_v<T>(ts...))>(a,make_v<T>(ts...)); }
template <typename T, typename V> typename enable_if<is_class<T>::value == 0>::type fill_v(T &t, const V &v) { t = v; }
template <typename T, typename V> typename enable_if<is_class<T>::value != 0>::type fill_v(T &t, const V &v) { for(auto &e : t) fill_v(e,v); }
const int MOD = 1000000007;
const int INF = 0x3f3f3f3f3f3f3f3f;
const double EPS = 1e-10;

signed main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(10);

  string s;
  cin >> s;
  int ans = s.size();
  REP(i,s.size()-1) if (s[i] != s[i+1]) CHMIN(ans,max(i+1,(int)s.size()-i-1));
  cout << ans << endl;

  return 0;
}