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 {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		String[] s = new String[n];
		long ans = 0;
		long startWithB = 0;
		long endWithA = 0;
		long both = 0;
		for (int i = 0; i < n; i++) {
			s[i] = sc.next();
			boolean isPrevA = false;
			for (int j = 0; j < s[i].length(); j++) {
				if (isPrevA && s[i].charAt(j) == 'B') {
					ans++;
				}
				if (s[i].charAt(j) == 'A')
					isPrevA = true;
				else
					isPrevA = false;
			}
			if (s[i].charAt(0) == 'B')
				startWithB++;
			if (s[i].charAt(s[i].length() - 1) == 'A')
				endWithA++;
			if ((s[i].charAt(0) == 'B') && (s[i].charAt(s[i].length() - 1) == 'A'))
				both++;
		}
		sc.close();
		if (!((both == startWithB) && (both == endWithA)))
			ans += Math.min(Math.min(startWithB, endWithA), n - 1);
		else if (both == 0)
			ans += Math.min(Math.min(startWithB, endWithA), 0);
		else
			ans += Math.min(Math.min(startWithB, endWithA), both - 1);
		System.out.println(ans);
	}
}