CF div2 334 C

C. Alternative Thinking

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 anot-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.

Sample test(s)

input

8
10000011

output

5

input

2
01

output

2

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.

记录00 11 这样的个数 如果有 ans++,如果大于等于2个 ans+=2 然后如果有000 这样的情况 ans+=2

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;

char s[100500];
int main()
{

    int n,k,ans=0,cnt=1,i;
    scanf("%d",&n);
    scanf("%s",s);
    for (i=1;i<n;i++)
    {
        if (s[i]==s[i-1]) ans++;
    }
    char tmp=s[0];
    for (i=1;i<n;i++)
    {
        if (s[i]!=tmp)
         {
            cnt++;
            tmp=s[i];
         }
    }

    ans=min(ans,2);
    ans += cnt;
    printf("%d
",ans);


}