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: Wrong Answer)

use std::cmp::{max, min};
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };

    ($next:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}
const MOVES: [(i32, i32); 4] = [(1, 0), (0, 1), (-1, 0), (0, -1)];
use std::collections::BinaryHeap;
fn main() {
    input!{
        n:i64,
        sn:[String;n],
    }
    let mut end_a = vec![];
    let mut start_b = vec![];
    let mut all = vec![];
    let mut ans = 0;
    for j in 0..sn.len() {
        if sn[j].starts_with("B") && sn[j].ends_with("A"){
            all.push(j);
        } else if sn[j].ends_with("A") {
            end_a.push(j);
        } else if sn[j].starts_with("B"){
            start_b.push(j);
        }
        if sn[j].contains("AB"){
            ans += 1;
        }
    }
    let mut alen = end_a.len();
    let mut blen = start_b.len();
    let mut can_a = true;
    let mut can_b = true;
    for j in 0..all.len() {
        if alen >0 {
            ans += 1;
            alen -=1;
        } else {
            can_a = false;
        }
        if blen > 0 {
            ans += 1;
            blen -=1;
        } else {
            can_b = false;
        }
        if (!can_a || !can_b) && j != 0 {
            ans += 1;
        }
    }
    if alen != 0 || blen != 0 {
        ans += min(alen,blen);
    }
    // println!("{:?}",all);
    // println!("{:?}",end_a );
    // println!("{:?}",start_b);
    println!("{}",ans);
}