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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define REP(i, n) for(int i = 0; i < n; i++)
#define ALL(v) (v).begin(), (v).end()

using namespace std;
using ll = long long;

int main()
{
  int N; cin >> N;
  vector<string> s(N); REP(i, N) cin >> s[i];

  ll ans = 0, acnt = 0, bcnt = 0, abcnt = 0;
  for (string ss : s) {
    for (int i = 1; i < ss.size(); ++i) {
      if (ss[i-1] == 'A' and ss[i] == 'B') ans++;
    }

    if (ss.front() == 'B' and ss.back() == 'A') abcnt++;
    else if (ss.front() == 'B') bcnt++;
    else if (ss.back() == 'A') acnt++;
  }

  // printf("ans: %lld, %lld, %lld, %lld\n", ans, acnt, bcnt, abcnt);

  cout << ans + min(acnt, bcnt)+max(abcnt-1, 0LL)+(abs(acnt-bcnt)>0?1:0) << endl;
}