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

#start_b_end_aとend_aとstart_bの組み合わせを何個作るか考える
start_b_and_a_set = [[end_a, start_b].min , 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 start_b_and_a_set < start_b_end_a
  extra_ab = start_b_end_a - start_b_and_a_set
else
  extra_ab = 0
end

ab_count += start_b_and_a_set * 2 + extra_ab

# ここまでで全てのstart_b_end_aが使い終わる
# 余ったstart_bとend_aで作れないか模索する
ab_count += [start_b - start_b_and_a_set, end_a -start_b_and_a_set].min

print("#{ab_count}\n")