Is there anything better than going to the zoo after a tiresome week at work? No wonder Grisha feels the same while spending the entire weekend accompanied by pretty striped zebras.
Inspired by this adventure and an accidentally found plasticine pack (represented as a sequence of black and white stripes), Grisha now wants to select several consequent (contiguous) pieces of alternating colors to create a zebra. Let’s call the number of selected pieces the length of the zebra.
Before assembling the zebra Grisha can make the following operation 0
or more times. He splits the sequence in some place into two parts, then reverses each of them and sticks them together again. For example, if Grisha has pieces in the order “bwbbw” (here ‘b’ denotes a black strip, and ‘w’ denotes a white strip), then he can split the sequence as bw|bbw (here the vertical bar represents the cut), reverse both parts and obtain “wbwbb”.
Determine the maximum possible length of the zebra that Grisha can produce.
Input
The only line contains a string s(1≤|s|≤105, where |s| denotes the length of the string s) comprised of lowercase English letters ‘b’ and ‘w’ only, where ‘w’ denotes a white piece and ‘b’ denotes a black piece.
Output
Print a single integer — the maximum possible zebra length.
Examples
Input
bwwwbwwbw
Output
5
Input
bwwbwwb
Output
3
Note
In the first example one of the possible sequence of operations is bwwwbww|bw →w|wbwwwbwb → wbwbwwwbw, that gives the answer equal to 5
.In the second example no operation can increase the answer.
這道題的點就是將字符串收尾連接起來即可
代碼：
//將收尾連接起來，找連續字符不相同的串的長度即可
#include<stdio.h>
#include<string>
#include<vector>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
string str;
int main()
{
cin>>str;
int len=str.size();
if(len==1)
{
printf("1\n");
return 0;
}
int ans=1,k=1;
for(int i=1;i<3*len;i++)
{
if(str[(i-1)%len]!=str[i%len])//循環3輪,數組會越界,所以注意要模長度
k++;
else
k=1;
ans=max(k,ans);
}
ans=min(ans,len);//由於循環三輪，所以ans可能會超過數組本身長度
printf("%d\n",ans);
//system("pause");
return 0;
}