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 min(int a, int b){
    return (a<b)?a:b;
}

int main(void){
    int N;
    scanf("%d", &N);
    int head = 0;
    int tail = 0;
    int head_tail = 0;
    int count = 0;
    for(int i = 0; i < N; i++){
        char s[11];
        scanf("%s", s);
        if(s[0] == 'B') head += 1;
        if(s[strlen(s)-1] == 'A') tail += 1;
        if(s[0] == 'B' && s[strlen(s)-1]== 'A') head_tail += 1;
        for (int j = 0; j < strlen(s)-1; j++){
            if (s[j] == 'A' && s[j + 1] == 'B'){
                count += 1;
            }
        }
    }
    if(head_tail > 0) {
        count += head_tail - 1;
        if(tail - head_tail > 0) count += 1;
        if(head - head_tail > 0) count += 1;
        if((tail - head_tail - 1 + tail - head_tail - 1) > 0){
            count += min(tail - head_tail - 1, tail - head_tail - 1);
        }
    }
    else if(head_tail == 0){
        count += min(head, tail);
    }
    printf("%d\n", count);
    return 0; 
}