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: Accepted)

N=int(input())

naiv=0
endsA=0
startsB=0
both=0

for i in range(N):
    s=input()
    naiv+=s.count('AB')
    if s[0] == 'B' and s[-1] == 'A':
        both += 1
    elif s[0] == 'B':
        startsB += 1
    elif s[-1] == 'A':
        endsA += 1

# print("endsA {}".format(endsA))
# print("startsB {}".format(startsB))
# print("both {}".format(both))
# print("naiv {}".format(naiv))

res=naiv
if both >= startsB and both >= endsA and (endsA > 0 and startsB > 0):
    lower = min(endsA, startsB)
    higher = max(endsA, startsB)
    res += lower*2
    both -= lower
    startsB -= lower
    endsA -= lower
    res += both
    both = 0
elif both <= startsB and both <= endsA:
    res += both*2
    endsA -= both
    startsB -= both
    both = 0
    res += min(endsA, startsB)
# まんなか
elif both > 0 and max(endsA, startsB) > 0:
    lower = min(endsA, startsB)
    res += 2*lower
    both -= lower
    endsA -= lower
    startsB -= lower
    res += both
elif both > 0 and max(endsA, startsB) == 0:
    res += both-1
else:
    res += min(endsA, startsB)

print(res)