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 <iostream>
#include <string>
using namespace std;
int main(){
  int N; cin >> N;
  //vector<string> s;
  int a=0, b=0, ans=0;
  //string AB = "AB"
  //int size = AB.size();
  for(int i=0; i<N; i++) {
    string s_hold;
    cin >> s_hold;
    //s.push_back(s_hold);
    if (s_hold[0] == 'B') {
      b++;
    }
    for (int p = 0; p < s_hold.length() - 1; p++) {
      if (s_hold[p] == 'A') {
        if (s_hold[p+1] == 'B') {
          ans++;
        }
      }
    }
    if (s_hold[s_hold.length() - 1] == 'A') {
      a++;
    }
    //int p;
    /*while (p != string::npos) {
       pos = s_hold.find(AB, pos + size);
       }*/
    //s_hold.find(AB)
  }

  if (a==N && b==N) {
    ans = ans + N - 1;
  }
  else {
    ans = ans + min(a, b);
  }
  cout << ans << endl;
  return 0;
}