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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>

int count_A(const char cA[])
{
	int A = 0;
	int L = 0;
	while (cA[L]) {
		L++;
	}
	if (cA[L - 1] == 'A')
		A = 1;
	return A;
}

int count_B(const char cB[])
{
	int B = 0;
	if (cB[0] == 'B')
		B = 1;
	return B;
}


int count_AB(const char cAB[])
{
	int i = 0, count = 0;
	while (cAB[i]) {
		if (cAB[i] == 'A' && cAB[i + 1] == 'B')
			count++;
		i++;
	}
	return count;
}

int main(void)
{
	char s[12];
	int A = 0, B = 0;
	int N, L = 0, count = 0;
	scanf("%d", &N);
	if (N == 1) {
		scanf("%s", s);
		count += count_AB(s);
	}
	else {
		for (int i = 0; i < N; i++) {
			scanf("%s", s);
			A += count_A(s);
			B += count_B(s);
			count += count_AB(s);
		}
	}
	if (A == B)
		count += A;
	else if (A < B)
		count += A;
	else
		count += B;
	printf("%d", count);
}