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

a_flag_count = 0
b_flag_count = 0
a_b_flag_count = 0
count = 0
n.times do |t|
  str = gets.chomp
  count += str.scan(/AB/).size
  a_flag = str[-1] == "A"
  b_flag = str[0] == "B"
  if a_flag && b_flag then
    a_b_flag_count += 1
  elsif a_flag then
    a_flag_count += 1
  elsif b_flag then
    b_flag_count += 1
  end
end

if b_flag_count > a_flag_count then
  tmp = a_flag_count
  a_flag_count = b_flag_count
  b_flag_count = tmp
end

diff = a_flag_count - b_flag_count
if diff == 0 then
  count += (a_flag_count + (a_b_flag_count - 1))
elsif a_b_flag_count >= diff then
  count += (a_flag_count + (a_b_flag_count - diff))
else
  count += (b_flag_count + a_b_flag_count)
end

puts count