Alternative Thinking

Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length n. Each character of Kevin's string represents Kevin's score on one of the n questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0, 1, 0, 1}, {1, 0, 1}, and {1, 0, 1, 0} are alternating sequences, while {1, 0, 0} and {0, 1, 0, 1, 1} are not.Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have.

Input

The first line contains the number of questions on the olympiad n (1 ≤ n ≤ 100 000).The following line contains a binary string of length n representing Kevin's results on the USAICO.

Output

Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring.

Example
Input

810000011Output5Input201Output2

Note

In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'.In the second sample, Kevin can flip the entire string and still have the same score.

#include<stdio.h>//转载声明：http://blog.csdn.net/lizhaowei213/article/details/50557390//感觉看完了解析，感觉还好啊，好像并不是这么难做，//只是自己一想到模拟怎么找到这个反转的起点和终点,确实是有点愁人//然而根本就不用去模拟，直接去想情况//说说这个增加的个数吧，如果我们增加的话，就是说明这段区间的两端，还有相同的//假设这段区间是101，也就是说明。。。11011。。。，这种情况，看到之前就是3，改成010之后就变成了5const int Max=1e5+1;char str[Max];int main(){   int n,cnt=0;   scanf("%d",&n);   scanf("%s",str);   if(n==1)    printf("1\n");   else   {     for(int i=1;i<n;i++)     {        if(str[i]!=str[i-1])            cnt++;     }     cnt++;     if(cnt==n)    //自己本身就是1010串，这样的话，就不用反转了，没有什么卵用        printf("%d\n",n);     else if(cnt==n-1)//就是说明不是连续的1010串，而且只有一个二连续的点，这样的话，就是反转后面的，直接就可以了        printf("%d\n",n);     else     {         printf("%d\n",cnt+2);//对于有多个连续区间的，或者一个区间过长的，我们也就只能反转一个连续的区间，最多也就只能增加2,他就要这个最大的     }   }   return 0;}