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 (1.42.0) (judge verdict: Accepted)

#[allow(unused_imports)]
use itertools::Itertools;
use proconio::input;
#[allow(unused_imports)]
use proconio::marker::*;

fn main() {
    input! {
        n: usize,
        s: [Chars; n],
    }

    let cnt = s
        .iter()
        .map(|s| s.windows(2).filter(|w| w == &['A', 'B']).count())
        .sum::<usize>();

    let mut both = 0usize;
    let mut first = 0;
    let mut last = 0;
    for s in s {
        let (f, l) = (s.first() == Some(&'B'), s.last() == Some(&'A'));
        if f && l {
            both += 1;
        } else if f {
            first += 1;
        } else if l {
            last += 1;
        }
    }
    use std::cmp::min;
    let mut res = cnt + both.saturating_sub(1);
    if both != 0 && first != 0 {
        res += 1;
    } else if both != 0 && last != 0 {
        res += 1;
    }

    println!("{}", res + min(first, last));
}