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>

#ifndef LOCAL_
#define fprintf if( false ) fprintf
#endif // LOCAL_
#define dump() fprintf(stderr, "#%s.%d\n", __func__, __LINE__);
#define dumpl(x1) fprintf(stderr, "#%s.%d (%s) = (%ld)\n", __func__, __LINE__, #x1, x1);
#define dumpll(x1, x2) fprintf(stderr, "#%s.%d (%s, %s) = (%ld, %ld)\n", __func__, __LINE__, #x1, #x2, x1, x2);
#define dumplll(x1, x2, x3) fprintf(stderr, "#%s.%d (%s, %s, %s) = (%ld, %ld, %ld)\n", __func__, __LINE__, #x1, #x2, #x3, x1, x2, x3);
#define dumpd(x1) fprintf(stderr, "#%s.%d (%s) = (%lf)\n", __func__, __LINE__, #x1, x1);
#define dumpdd(x1, x2) fprintf(stderr, "#%s.%d (%s, %s) = (%lf, %lf)\n", __func__, __LINE__, #x1, #x2, x1, x2);
#define loop for(;;)
#define guard(x) if( not ( x ) ) continue;
typedef int64_t Int;
typedef std::vector<Int> LI;
const double pi = 3.14159265358979323846264338;
const Int mod = 1000000007;

template<typename T> void scan1(T& x) { fprintf(stderr, "unknown type\n"); }
template<> void scan1(Int& x) { if( scanf("%" PRId64, &x) < 0 ) exit(0); }
template<> void scan1(std::string& x) { if( not ( std::cin >> x ) ) exit(0); }
template<> void scan1(LI& xs) { for(Int &x : xs) scan1(x); }
void scan() {}
template<typename Head, typename... Tail>
void scan(Head& x, Tail&... xs) {
  scan1(x); scan(xs...);
}

template<typename T>
struct Maybe {
   T x;
   bool valid;
   Maybe() : valid(false) {}
   Maybe(T x_) : x(x_), valid(true) {}
   operator T() const {
      assert( valid );
      return x;
   }
};

struct Solver {
   Solver() { fprintf(stderr, "--------Solver begin--------\n"); }
   ~Solver() { fprintf(stderr, "--------Solver end--------\n"); }
   void solve() {
      std::string ss; scan(ss);
      Int n = ss.size();
      Int t = 0;
      if( n % 2 == 1 ) t += 1;
      for(Int i = n / 2 - 1; i >= 0; --i) {
         if( ss[i] == ss[n-i-1] and ss[i] == ss[n/2] ) t += 2;
         else break;
      }
      dumpl(t);
      Int res1 = (n + t) / 2;
      // Int res2 = -1;
      // for(Int i = 1; i < n; ++i) {
      //    if( ss[i-1] != ss[i] ) {
      //       res2 = i;
      //       break;
      //    }
      // }
      // Int res3 = -1;
      // for(Int i = n - 1; i >= 1; --i) {
      //    if( ss[i-1] != ss[i] ) {
      //       res3 = n - i;
      //    }
      // }
      // Int res = std::max(std::max(res1, res2), res3);
      Int res = res1;
      printf("%ld\n", res);
   }
};

int main() {
   loop std::unique_ptr<Solver>(new Solver())->solve();
}