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 inputs = Array(n) { readLine()!!}

    // 最初から文字列内に存在する AB の数
    val initialTotal = inputs.sumBy { countAB(it) }


    // 結合することで新たに作りうる AB の数
    var aEnd = 0
    var bStart = 0
    var aEndBStart = 0

    for (s in inputs){
        if (s[0] == 'B' && s[s.lastIndex] == 'A') {
            aEndBStart += 1
        } else {
            if (s[0] == 'B'){
                bStart += 1
            }
            if (s[s.lastIndex] == 'A'){
                aEnd += 1
            }
        }
    }

    val pair = Math.min(aEnd,bStart)
    val rest = Math.abs(aEnd-bStart)

    if (rest >= aEndBStart) {
        print(initialTotal + pair + rest)
    } else {
        if (rest == 0){
            print(initialTotal + pair + aEndBStart - 1)
        } else {
            print(initialTotal + pair + aEndBStart)
        }

    }
}


fun countAB(s: String): Int {
    var cnt = 0
    for (i in 0 until s.length - 1) {
        if(s[i] == 'A' && s[i+1] == 'B') {
            cnt += 1
        }
    }

    return cnt
}