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 Scala (2.11.7) (judge verdict: Wrong Answer)

import java.util.Scanner

import scala.annotation.tailrec
import scala.collection.mutable.ArrayBuffer


object Main {
  def read() = {
    val sc = new Scanner(System.in)
    val n = sc.nextInt()
    val strs = for (_ <- 0 until n) yield sc.next()
    (n, strs)
  }

  def solve(n:Int, strs: IndexedSeq[String]): Int={
    // AB含み
    // A終わり
    // B始まり
    // A終わりB始まり

    val nContainsAB = strs.count(str => str.contains("AB"))
    val nAEnd = strs.count(str => str.last == 'A')
    val nBBegin = strs.count(str => str.head == 'B')
    val nAEndBBEgin = strs.count(str => str.last == 'A' && str.head=='B')

    if(nAEndBBEgin == nAEnd && nAEndBBEgin == nBBegin){
      nContainsAB + nAEndBBEgin - 1
    }else{
      nContainsAB + Math.min(nAEnd, nBBegin)
    }
  }

  def main(args: Array[String]): Unit = {
    val (n, strs) = read()
    println(solve(n,strs))
  }
}