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 Ruby (2.3.3) (judge verdict: Wrong Answer)

n = gets.chomp.to_i
start_b_strings= []
end_a_strings = []
start_b_end_a_strings = []
ab_count = 0

for i in 0..n-1
  words = gets.chomp
  if !words.scan("AB").empty?
    ab_count += words.scan("AB").length
  end
  if words[0] == "B" && words[-1] == "A"
    start_b_end_a_strings.push(words)
    next
  end
  start_b_strings.push(words) if words[0] == "B"
  end_a_strings.push(words) if words[-1] == "A"
end

start_b = start_b_strings.length
end_a = end_a_strings.length
start_b_end_a = start_b_end_a_strings.length
if start_b_end_a < end_a &&
  start_b_end_a < start_b

  # start_b_end_aを真ん中に挟んでABを二つずつ作ることができる
  ab_count += start_b_end_a * 2
  # それ以外はそれぞれカウントしていく
  ab_count += [start_b - start_b_end_a, end_a - start_b_end_a].min
else
  # start_b_end_aが余る場合
  # 1.start_b_end_aを真ん中に挟んでABを二つずつできるだけ作る
  start_b_end_a_set1 = [start_b, end_a].sort[0]
  ab_count += start_b_end_a_set1 * 2
  # 2.時にはstart_b_end_aを二つ使う
  start_b_end_a_set2 =  [[start_b, end_a].sort[1] - start_b_end_a_set1, (start_b_end_a - start_b_end_a_set1) / 2].min
  ab_count += start_b_end_a_set2 * 2

  # 余ったのでなんか作る
  amari_a_or_b = [start_b, end_a].sort[1] - start_b_end_a_set1 - start_b_end_a_set2
  amari_a_and_b = start_b_end_a - start_b_end_a_set1 - (start_b_end_a_set2 * 2)
  ab_count += amari_a_or_b > amari_a_and_b ? amari_a_and_b : amari_a_or_b
end
print("#{ab_count}\n")