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 _USE_MATH_DEFINES

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <string>
#include <utility>
#include <climits>
#include <vector>
#include <map>
#include <queue>
#include <set>
using namespace std;

#define ll long long
const int MOD = 1000000007;//10^9+7
const string SPACE = " ";

ll powsur(ll a, ll b, ll p) {

	if (b == 0) {

		return 1;

	}
	else if (b % 2 == 0) {

		ll d = powsur(a, b / 2, p);
		return ((d%p)*(d%p)) % p;

	}
	else {

		return ((a%p)*powsur(a, b - 1, p) % p) % p;

	}

}
//pは素数,aのb乗をpで割った余り

ll CMOD(ll n, ll r, ll p) {

	ll a = 1, b = 1, c = 1, re;
	for (int i = 1; i <= n; i++) {

		a *= i;
		a = a%p;

	}
	for (int i = 1; i <= r; i++) {

		b *= i;
		b = b%p;

	}
	for (int i = 1; i <= n - r; i++) {

		c *= i;
		c = c%p;

	}
	b = powsur(b, p - 2, p);
	c = powsur(c, p - 2, p);
	re = ((a%p)*(b%p)) % p;
	re = ((re%p)*c%p) % p;

	return re;

}
//pは素数,(nCr)%p

ll gcd(ll a, ll b) {

	if (a > b) {

		ll tmp = a;
		a = b;
		b = tmp;

	}
	ll r;
	r = b%a;
	if (r == 0)return a;
	else return gcd(r, a);

}
//最大公約数

ll lcm(ll a, ll b) {

	return (a / gcd(a, b))*b;

}
//最小公倍数

ll MAX(ll a, ll b) {

	if (a > b)return a;
	else return b;

}
//MAX

ll MIN(ll a, ll b) {

	if (a < b)return a;
	else return b;

}
//MIN

bool func(ll p) {

	if (p == 1)return false;
	else if (p == 2)return true;
	else {

		bool flag = true;
		for (int i = 2; i <= sqrt(p); i++) {

			if (p%i == 0) {

				flag = false;
				break;

			}

		}
		return flag;

	}

}
//素数判定,素数=true,合成数=false

int main() {

	string S;
	ll ans = 1000000, tmp;

	cin >> S;
	for (ll i = 0; i < S.size() - 1; i++) {

		if (S[i] != S[i + 1])tmp = max((ll)(i + 1), (ll)(S.size() - i - 1));
		ans = MIN(ans, tmp);

	}
	cout << ans << endl;

	return 0;
}