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

# end_aとstart_bの組み合わせを何個作るか考える
start_b_and_end_a_set = [end_a, start_b].min

if 0 < start_b_and_end_a_set
  # end_a | start_b_and_end_a | start_bの組み合わせがコスパ良い
  best_set = [start_b_and_end_a_set, start_b_end_a].min

  # もしstart_b_end_aが余るようであれば真ん中に全部挟んでABをたくさん作ることができる
  if 0 < start_b_end_a - best_set
    extra_ab = start_b_end_a - best_set
  else
    extra_ab = 0
  end
  ab_count += best_set + start_b_and_end_a_set + extra_ab
else
  # end_a とstart_b_and_end_aまたは start_b_and_end_aとstart_bで組み合わせを作っていく
  best_set = [[end_a, start_b].max, start_b_end_a].min
  # もしstart_b_end_aが余るようであればどこかに繋げとけばABをたくさん作ることができる
  if 0 < start_b_end_a - best_set
    extra_ab = start_b_end_a - best_set
  else
    extra_ab = 0
  end
  ab_count += best_set + extra_ab
end

print("#{ab_count}\n")