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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

import java.util.Arrays
import java.util.ArrayList

const val MOD: Long = (1e9+7).toLong()
fun Double.format(digits: Int) = String.format("%.${digits}f", this)

fun main(args: Array<String>) {
    val n = readLine()!!.toInt()
    val s = Array(n, {readLine()!!})

    var cnt = 0
    var cnta = 0
    var cntb = 0
    var non = 0
    for (t in s) {
        for (i in 0..t.length-2) {
            cnt += if (t.slice(i..i+1) == "AB") 1 else 0
        }
        cnta += if (t.last() == 'A') 1 else 0
        cntb += if (t[0] == 'B') 1 else 0
        non += if (t.last() != 'A' && t[0] != 'B') 1 else 0
    }

    if (cnta == n-non && cntb == n-non) {
        println(cnt+cnta-1)
    } else {
        println(cnt + Math.min(cnta, cntb))
    }
}