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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdint.h>
#include <inttypes.h>

int main (int argc, char const *argv[])
{
    int64_t N;
    char s[(int)1e4][11];
    int64_t initial_b = 0, final_a = 0, both = 0, ans = 0, flag = 0;

    scanf("%"PRId64, &N);
    for (int i = 0; i < N; i++) {
        int p = 0;
        scanf("%s", s[i]);
        while (s[i][p]) {
            if ((s[i][p] == 'A') && (s[i][p + 1] == 'B')) ans++;
            p++;
        }
        if ((s[i][p - 1] == 'A') && (s[i][0] == 'B')) both++;
        else if (s[i][0] == 'B') initial_b++;
        else if (s[i][p - 1] == 'A') final_a++;
    }

    ans += initial_b > final_a ? final_a : initial_b;
    if ((initial_b != 0) || (final_a != 0)) {
        ans += both;
    } else {
        ans += both - 1;
    }

    printf("%"PRId64"\n", ans);
}