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: Accepted)

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

int count_AB(char[], int n);


int main(void)
{
	char s[12];
	int A = 0, B = 0, AB = 0;
	int N, L = 0,C = 0;
	scanf("%d", &N);
	for (int i = 0; i < N; i++) {
		scanf("%s", s);
		L = strlen(s);
		C += count_AB(s, L);
		if (s[0] == 'B' && s[L - 1] == 'A') {
			AB++;
			A++;
			B++;
		}
		else if (s[0] == 'B' && s[L - 1] != 'A') {
			B++;
		}
		else if (s[0] != 'B' && s[L - 1] == 'A') {
			A++;
		}
		else {
			continue;
		}
	}
	if (AB == B && B == A && A != 0) {
		printf("%d", C += AB - 1);
	}
	else if (A > B) {
		printf("%d", C += B);
	}
	else {
		printf("%d", C += A);
	}
	return 0;
}

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