/*
n%3=0，整个数组直接和PDD比较
n%3=1，枚举不是pdd的位置，过一遍前缀和
n%3=2，枚举两个不是pdd的位置，过一遍前缀和
*/
#include <bits/stdc++.h>

using namespace std;

const int N = 1e4 + 10;
int f[N], n;
string s;

int main()
{
    cin >> n >> s;
    s = " " + s;
    for (int i = 0; i <= 2; i ++ )
        for (int j = i + 3; j <= n; j += 3)
            f[j] = f[j - 3] + abs(s[j - 2] - 'P') + abs(s[j - 1] - 'D')
                            + abs(s[j] - 'D');  

    int ans = 1e9;               
    if (n % 3 == 0 || n < 3) ans = f[n] - f[0];
    else if (n % 3 == 1)
        for (int i = 1; i <= n; i += 3 )
            ans = min(ans, f[i - 1] - f[0] + f[n] - f[i - 1])
    else if (n % 3 == 2)
        for (int i = 1; i <= n; i += 3)
            for (int j = i + 4; j <= n; j += 3)
                ans = min(ans, f[j - 1] - f[i] + f[n] - f[j]);

    cout << ans;
    return 0;

}