Alternative Thinking 找规律

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

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.

懂了意思就好了 ，只允许该连续的子串，不可以跳着改，如果有三个连续的，原来的数量加2，比如10001，原来是3，改了之后为10101，变成5，如果有两个连续的，加1，比如1101，原来是3，改了之后0101，变为4，加入有间隔的两个及以上这种模式，那么原数量加2，如11011，原来是3，改了之后10101，变成5，下划线代表子串被改变。

代码：

#include <iostream>
#include <cstdio>
#include <queue>
#include <map>
#include <algorithm>
using namespace std;

int main()
{
    int n,c=,er=,san=,min=;
    char str[]="";
    scanf("%d",&n);
    getchar();
    for(int i=;i<=n+;i++)
    {
        if(i<n+)scanf("%c",&str[i]);
        if(str[i]!=str[i-])
        {
            if(c==)er++;
            else if(c>=)san++;
            if(i<n+)min++;
            c=;
        }
        else c++;
    }
    //cout<<er<<' '<<san<<endl;
    if(san||er>)min+=;
    else if(er) min+=;
    printf("%d",min);
}