Problem:
Score : 400 points

Problem Statement

Snuke has N strings. The i-th string is s_i.

Let us concatenate these strings into one string after arranging them in some order.
Find the maximum possible number of occurrences of AB in the resulting string.

Constraints

1 \leq N \leq 10^{4}

2 \leq |s_i| \leq 10

s_i consists of uppercase English letters.

Input

Input is given from Standard Input in the following format:

N
s_1
\vdots
s_N

Output

Print the answer.

Sample Input 1

3
ABCA
XBAZ
BAD

Sample Output 1

2

For example, if we concatenate ABCA, BAD and XBAZ in this order, the resulting string ABCABADXBAZ has two occurrences of AB.

Sample Input 2

9
BEWPVCRWH
ZZNQYIJX
BAVREA
PA
HJMYITEOX
BCJHMRMNK
BP
QVFABZ
PRGKSPUNA

Sample Output 2

4

Sample Input 3

7
RABYBBE
JOZ
BMHQUVA
BPA
ISU
MCMABAOBHZ
SZMEHMA

Sample Output 3

4

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)


import java.util.*;

public class Main {

    void run() {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int ans = 0;
        LinkedList<Integer> as = new LinkedList<>(), bs = new LinkedList<>();
        HashSet<Integer> ain = new HashSet<>();

        for (int i = 0; i < n; i++) {
            String s = sc.next();
            for (int j = s.length() - 2; j > 0; j--) {
                if (s.charAt(j - 1) == 'A' && s.charAt(j) == 'B') ans++;
            }
            if (s.charAt(0) == 'B') {
                if (s.charAt(s.length() - 1) == 'A') bs.addFirst(i);
                else bs.addLast(i);
            }
            if (s.charAt(s.length() - 1) == 'A') {
                if (s.charAt(0) == 'B') {
                    as.addLast(i);
                } else {
                    as.addFirst(i);
                }
                ain.add(i);
            }
        }
//        debug(ans, as, bs);

        boolean[] used = new boolean[n];
        if (!as.isEmpty()) {
            int i = as.removeFirst();
            while (!bs.isEmpty()) {
                if (ain.contains(i)) {
                    // nothing
                } else {
                    used[i] = true;
                    while (!as.isEmpty() && used[i]) i = as.removeFirst();
                    if (used[i]) break;
                }
                used[i] = true;
                while (!bs.isEmpty()) {
                    int j = bs.remove();
                    if (used[j]) continue;
//                debug(i, j);
                    i = j;
                    ans++;
                    break;
                }
            }
        }
        System.out.println(ans);
    }

    void debug(Object...os) {
        System.err.println(Arrays.deepToString(os));
    }

    public static void main(String[] args) {
        new Main().run();
    }
}