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 Rust (1.15.1) (judge verdict: Accepted)

use std::io::*;

fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let n: usize = itr.next().unwrap().parse().unwrap();

    let mut ans = 0;
    let mut a = 0;
    let mut b = 0;
    let mut ba = 0;
    for _ in 0..n {
        let s: Vec<char> = itr.next().unwrap().chars().collect();
        for i in 0..s.len() - 1 {
            if s[i] == 'A' && s[i + 1] == 'B' {
                ans += 1;
            }
        }
        if s[0] == 'B' && s[s.len() - 1] == 'A' {
            ba += 1;
        } else if s[0] == 'B' {
            b += 1;
        } else if s[s.len() - 1] == 'A' {
            a += 1;
        }
    }
    if ba > 0 {
        ans += ba - 1;
        if a > 0 {
            a -= 1;
            ans += 1;
        }
        if b > 0 {
            b -= 1;
            ans += 1;
        }
    }
    ans += std::cmp::min(a, b);

    println!("{}", ans);
}