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)

fun main(args: Array<String>) {
    val n = readLine()!!.toInt()
    val sList = Array(n) { readLine()!! }
    val countAB = sList.map {
        var count = 0
        for (i in 0 until it.length - 1) {
            if (it[i] == 'A' && it[i + 1] == 'B') count++
        }
        count
    }.sum()
    val countBA = sList.count { it.first() == 'B' && it.last() == 'A' }
    val countLastA = sList.count { it.first() != 'B' && it.last() == 'A' }
    val countFirstB = sList.count { it.first() == 'B' && it.last() != 'A' }
    if (countBA > 0) {
        println(countAB + countBA - 1 + Math.min(countLastA, countFirstB) + if ((countLastA != 0 || countFirstB != 0) && countLastA != countFirstB) 1 else 0)
    } else if (countFirstB > 0 && countLastA > 0) {
        println(countAB + Math.min(countLastA, countFirstB))
    } else {
        println(countAB)
    }
}