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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

N = int(input())
ss = [input() for _ in range(N)]

contained = sum([s.count('AB') for s in ss])
B_hoge = sum([1 for s in ss if s[0]=='B'])
hoge_A = sum([1 for s in ss if s[-1]=='A'])
B_hoge_A = sum([1 for s in ss if s[0]=='B' and s[-1]=='A'])

B_hoge -= B_hoge_A
hoge_A -= B_hoge_A
gap = abs(B_hoge - hoge_A)
# 差がB_hoge_Aより大きい場合、可能な限り差を埋める
ans = contained
if gap >= B_hoge_A:
    ans += min(B_hoge,hoge_A) + B_hoge_A
#　差がB_hoge_Aより小さい場合、差を埋めた残りでペアを作る
else:
    ans += min(B_hoge,hoge_A)
    if B_hoge_A > 0:
        ans += B_hoge_A - 1
        if B_hoge!=hoge_A:
            ans += 1
print(ans)

# print('-'*20)
# print('contained:',contained)
# print(B_hoge, hoge_A, B_hoge_A)