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: Accepted)

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

int main()
{
    int numOfStrings;
    scanf("%d", &numOfStrings);

    int numBxA = 0, numBeginB = 0, numEndA = 0, numAB = 0;
    char str[11] = {0};
    for (int i = 0; i < numOfStrings; i++)
    {
        scanf("%s", str);
        int len = strlen(str);

        if (str[0] == 'B' && str[len - 1] == 'A') // 'B' ではじまり 'A' で終わる文字列を考慮する
            numBxA++;
        else if (str[0] == 'B')
                numBeginB++;
        else if (str[len - 1] == 'A')
            numEndA++;

        for (int j = 0; j < len - 1; j++)
        {
            if (str[j] == 'A' && str[j + 1] == 'B')
                numAB++;
        }
    }
    if (numBeginB + numEndA != 0)
        numAB += numBxA + ((numBeginB < numEndA) ? numBeginB : numEndA);
    else if (numBxA != 0)
        numAB += numBxA - 1;

    printf("%d\n", numAB);
    return 0;
}