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: Runtime Error)

def solve():
    N = int(input())
    words = ["a"]*N
    for i in range(N):
        words[i] = list(input())
    A_list = []
    B_list = []
    AB_list = [] 
    AB_count = 0
    for i in range(N):
        tag_A = tag_B = tag_AB = 0
        for j in range(len(words[i])):
            if words[i][j] == "A" and j == len(words[i])-1:
                tag_A = 1
                tag_AB = 0
            elif words[i][j] == "A":
                tag_AB = 1
            elif words[i][j] == "B" and tag_AB == 1:
                tag_AB = 0
                AB_count += 1
            elif words[i][j] == "B" and j == 0:
                tag_B = 1
                tag_AB = 0
            else:
                tag_AB = 0
        if tag_A == 1 and tag_B == 1:
            AB_list.append(i)
        elif tag_A == 1:
            A_list.append(i)
        elif tag_B == 1:
            B_list.append(i)
        #print(tag_A,tag_B,AB_count,A_list,B_list,AB_list)

    if len(AB_list) > 0 and len(A_list) > 0 and len(B_list) > 0:
        count = AB_count + len(AB_list) + min(len(A_list),len(B_list))
    elif len(AB_list) > 0 and len(A_list) == 0 and len(B_list) == 0:
        count = AB_count + len(AB_list) -1
    elif len(AB_list) > 0 and (len(A_list) == 0 or len(B_list) == 0):
        count = AB_count + len(AB_list)
    else:
        count = AB_count + min(len(A_list),len(B_list))

    print(count)


 solve()