/*
ID: wuqi9395@126.com
PROG: beads
LANG: C++
*/

// O(n^2)
#include<iostream>
#include<fstream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<ctype.h>
#include<algorithm>
#include<string>
#define PI acos(-1.0)
#define maxn 1000
#define INF 1<<25
#define mem(a, b) memset(a, b, sizeof(a))
typedef long long ll;
using namespace std;
int main ()
{
    string str;
    int n, i, l, r, mx = 0;
    ifstream fin ("beads.in");
    ofstream fout ("beads.out");
    fin>>n;
    fin>>str;
    str += str;
    for (i = 0; i < n; i++)
    {
        l = r = 1;
        int k = i + 1;
        char flag;
        while(l < n)
        {
            int j = i;
            while(str[j] == ‘w‘) j++;
            flag = str[j];
            if (str[k] == flag || str[k] == ‘w‘) k++, l++;
            else break;
        }
        k = i + n - 2;
        while(r < n)
        {
            int j = i;
            while(str[j] == ‘w‘) j--;
            flag = str[j];
            if (str[k] == str[i + n - 1] || str[k] == ‘w‘) k--, r++;
            else break;
        }
        if (l + r > mx) mx = l + r;
    }
    if (mx > n) mx = n;
    fout<<mx<<endl;
    return 0;
}

r[0] = p[0] = 0
 If c = ‘r‘ then r[p+1] = r[p] + 1 and b[p+1] = 0
        because the length of the blue beads is 0.
 if c = ‘b‘ then b[p+1] = b[p] + 1 and r[p+1] = 0
 if c = ‘w‘ then both length of the red and length of blue beads
             can be longer.
so r[p+1] = r[p]+1 and b[p+1] = b[p] + 1.

/*
ID: wuqi9395@126.com
PROG: beads
LANG: C++
*/
// O(N) DP
#include<iostream>
#include<fstream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<ctype.h>
#include<algorithm>
#include<string>
#define PI acos(-1.0)
#define maxn 1000
#define INF 1<<25
#define mem(a, b) memset(a, b, sizeof(a))
typedef long long ll;
using namespace std;
int main ()
{
    ifstream fin ("beads.in");
    ofstream fout ("beads.out");
    int i, j, n, mx = 0;
    int left[810][2] = {0}, right[810][2] = {0};
    string str;
    fin>>n>>str;
    str += str;
    for (i = 1; i <= 2 * n; i++)
    {
        if (str[i - 1] == ‘r‘)
            left[i][0] = left[i - 1][0] + 1, left[i][1] = 0;
        else if (str[i - 1] == ‘b‘)
            left[i][1] = left[i - 1][1] + 1, left[i][0] = 0;
        else
            left[i][1] = left[i - 1][1] + 1, left[i][0] = left[i - 1][0] + 1;
    }
    for (i = 2 * n - 1; i >= 0; i--)
    {
        if (str[i] == ‘r‘)
            right[i][0] = right[i + 1][0] + 1, right[i][1] = 0;
        else if (str[i] == ‘b‘)
            right[i][1] = right[i + 1][1] + 1, right[i][0] = 0;
        else
            right[i][1] = right[i + 1][1] + 1, right[i][0] = right[i + 1][0] + 1;

    }
    for (i = 0; i < 2 * n; i++)
        mx = max(mx, max(left[i][0], left[i][1]) + max(right[i][0], right[i][1]));
    mx = min(mx, n);
    fout<<mx<<endl;
    return 0;
}