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 (GCC 5.4.1) (judge verdict: Wrong Answer)


#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>

int compare_int(const void *a, const void *b)
{
    return *(int*)a - *(int*)b;
}

int main(void)
{
	//変数の宣言
	char s[100010];
	int len;
	int i,j;
	int flag=0;
	int ans;
	//データの読み込み
	scanf("%s",&s);
	
//	printf("nは%dです\n", n);
//	printf("データの読み込み終了\n");
	//実際の処理
	len=strlen(s);
	if(len%2==1){
		i=1;
		while(flag==0){
			if(s[len/2]!=s[len/2+i]||s[len/2]!=s[len/2-i]){
				i--;
				flag=1;
			}else{
				i++;
				if(2*i-1==len){
					i--;
					flag=1;
				}
			}
		}
		ans=len/2+i+1;
	}else{
		if(s[len/2-1]!=s[len/2]){
			ans=len/2;
		}else{
			i=1;
			if(s[len/2]!=s[len/2+i]||s[len/2]!=s[len/2-i-1]){
				flag=1;
			}else{
				i++;
				if(2*i==len+2){
					i--;
					flag=1;
				}
			}
			ans=len/2+i;
		}
	}
	
//	printf("計算部分終了\n");
	
	//出力
	printf("%d",ans);
	
//	printf("結果の出力終了\n");
	
}