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
texts = []
count = 0
right_hand = 0
left_hand = 0
double_hand = 0
N.times do |i|
    hands = [0,0]
    text = gets.chomp
    count += text.scan("AB").length
    text = text.split("").map(&:to_s)
    text = [text.first,"X",text.last]
    if text.last == "A"
        if text.first == "B"
            double_hand += 1
        else
            right_hand += 1
        end
    elsif text.first == "B"
        left_hand += 1
    end
end


left_hand += double_hand/2
right_hand += double_hand/2
count += double_hand/2
double_hand = double_hand%2


if left_hand >= right_hand
    count += right_hand
    left_hand = left_hand - right_hand
    if double_hand >= left_hand
        count += left_hand
        double_hand = double_hand - left_hand
        count += double_hand/2
    else
        count += double_hand
    end
else
    count += left_hand
    right_hand = right_hand - left_hand
    if double_hand >= right_hand
        count += right_hand
        double_hand = double_hand - right_hand
        count += double_hand/2
    else
        count += double_hand
    end
end
p count