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)

from collections import Counter
from collections import deque
from functools import reduce
from pprint import pprint
# import statistics
import bisect
import copy
import fractions
import math
import random
import sys
import time
sys.setrecursionlimit(10**7)
INF = 10 ** 18
MOD = 10 ** 9 + 7
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def MI(): return map(int, sys.stdin.readline().split())
def II(): return int(sys.stdin.readline())
def IS(): return input()
def C(x): return Counter(x)
def GCD_LIST(numbers):
    return reduce(fractions.gcd, numbers)
def LCM_LIST(numbers):
    return reduce(LCM, numbers)
def LCM(m, n):
    return (m * n // fractions.gcd(m, n))
def unite(x, y):
    # それぞれのノードの根を求める
    x = root(x)
    y = root(y)
    if x == y:
        return
    # node[x]の根をyに変更する
    node[x] = y
def same(x,y): return bool(root(x) == root(y))
def root(x):
    if node[x] == x:  # xが根の場合
        return x
    else:
        node[x] = root(node[x])  # 経路圧縮
        return node[x]
def dfs(v):
    # ノードに訪れた
    visited[v] = True
    for v2 in range(n):
        if graph[v][v2] == False:  # そもそもvとv2が辺でつながっていない
            continue
        if visited[v2] == True:  # もう見た
            continue
        dfs(v2)

n = II()
s = [IS() for _ in range(n)]
cnt = 0
sentou_b = 0
ushiro_a = 0
ab = 0
for i in s:
    cnt += i.count('AB')
    if i[0] == 'B':
        sentou_b += 1
    if i[-1] == 'A':
        ushiro_a += 1
    if i[0] == 'B' and i[-1] == 'A':
        ab += 1
# print(cnt)
# print(sentou_b)
# print(ushiro_a)
# print(ab)
# print('===')
if ab > 0 and sentou_b == ushiro_a:
    print(cnt+min(sentou_b, ushiro_a)-1)
else:
    print(cnt+min(sentou_b, ushiro_a))