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: Wrong Answer)

import java.util.Scanner;

class Solver {
	static final Solver INSTANCE = new Solver();

	void solve(Scanner sc) {
		int N = sc.nextInt();
		String[] S = new String[N];
		for (int i = 0; i < N; i++) {
			S[i] = sc.next();
		}

		int count = 0;
		int ba = 0;
		int startsWithB = 0;
		int endsWithA = 0;
		for (String s : S) {
			char[] chars = s.toCharArray();
			boolean sb = chars[0] == 'B';
			boolean ea = chars[chars.length - 1] == 'A';

			if (sb && ea){
				ba++;
			} else if (sb) {
				startsWithB++;
			} else if (ea) {
				endsWithA++;
			}

			for (int j = 1; j < chars.length; j++) {
				if (chars[j - 1] == 'A' && chars[j] == 'B') {
					count += 1;
				}
			}
		}
		count += ba / 2;
		count += endsWithA == startsWithB ? endsWithA : Math.min(endsWithA, startsWithB) + Math.min(ba, 1);
		System.out.println(count);
	}
}

class Main {
	public static void main(String... args) {
		Scanner in = new Scanner(System.in);

		Solver.INSTANCE.solve(in);

		in.close();
	}
}