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 LuaJIT (2.0.4) (judge verdict: Wrong Answer)

local ior = io.input()
local n = ior:read("*n", "*l")
local num_b_st_a_ed = 0
local num_b_st = 0
local num_a_ed = 0

local tot = 0

for i = 1, n do
  local str = ior:read()
  local len = #str
  local isfirst = true
  local bcand = false
  local adet = false
  if(str:sub(1, 1) == "B") then bcand = true end
  if(str:sub(len, len) == "A") then adet = true end
  for j = 1, len - 1 do
    if(str:sub(j, j) == "A" and str:sub(j + 1, j + 1) == "B") then
      tot = tot + 1
      -- print("FOUND")
    end
  end
  if(adet and bcand) then
    num_b_st_a_ed = num_b_st_a_ed + 1
    -- print(i, "AB")
  elseif(adet) then
    num_a_ed = num_a_ed + 1
    -- print(i, "A")
  elseif(bcand) then
    num_b_st = num_b_st + 1
    -- print(i, "B")
  end
end

local min = math.min(num_a_ed, math.min(num_b_st, num_b_st_a_ed))
tot = tot + min * 2
num_a_ed, num_b_st, num_b_st_a_ed = num_a_ed - min, num_b_st - min, num_b_st_a_ed - min

if(num_b_st_a_ed == 0) then
  tot = tot + math.min(num_a_ed, num_b_st)
elseif(num_a_ed == 0) then
  if(0 < num_b_st_a_ed) then
    tot = tot + (num_b_st_a_ed - 1)
    if(0 < num_b_st) then tot = tot + 1 end
  end
else
  if(0 < num_b_st_a_ed) then
    tot = tot + (num_b_st_a_ed - 1)
    if(0 < num_a_ed) then tot = tot + 1 end
  end
end
-- print(num_a_ed, num_b_st, num_b_st_a_ed)
print(tot)