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 Lua (5.3.2) (judge verdict: Accepted)

local counter=0
local BA,B_,_A=0,0,0
local n=io.read("n","l")
for i=1,n do
    local s=io.read()
    local _,sAB=s:gsub("AB","")
    counter=counter+sAB
    sf=s:sub(1,1)
    sl=s:sub(#s,#s)
    BA=((sf=="B" and sl=="A") and BA+1 or BA)
    B_=((sf=="B" and sl~="A") and B_+1 or B_)
    _A=((sf~="B" and sl=="A") and _A+1 or _A)
end

if BA==0 then
    counter=counter+math.min(B_,_A)
elseif BA>0 and B_+_A>0 then
    counter=counter+BA+math.min(B_,_A)
elseif BA>0 and B_+_A==0 then
    counter=counter+BA-1
end
print(counter)