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 Python (3.4.3) (judge verdict: Wrong Answer)

#https://atcoder.jp/contests/diverta2019/tasks/diverta2019_c

N = int(input())
s = []
for i in range(N):
    s.append(list(str(input())))

flag = False
count = 0
for i in range(len(s)):
    for j in range(len(s[i])):
        if s[i][j] == 'A':
            flag = True
        elif flag == True and s[i][j] == 'B':
            flag = False
            count += 1
        else:
            flag = False

startcnt = [0] * N
endcnt = [0] * N
for i in range(len(s)):
    if s[i][0] == 'B':
        startcnt[i] = 1
    if s[i][len(s[i])-1] == 'A':
        endcnt[i] = 1

cntpls = [0,0,0]
for i in range(N):
    if startcnt[i] + endcnt[i] == 1:
        if startcnt[i] == 1:
            cntpls[0]+= 1
        else:
            cntpls[1] += 1
    elif startcnt[i] + endcnt[i] == 2:
        cntpls[2] += 1

count += min(cntpls[0],cntpls[1]) + min(abs(cntpls[0]-cntpls[1]),cntpls[2])


print(count)