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 ( void ) {
    int N, ans = 0, _a = 0, b_ = 0, ba = 0;
    cin >> N;
    string s[N];
    for (int i = 0; i < N; i++) {
        cin >> s[i];
        int length = s[i].length();
        if (s[i][0] == 'B' && s[i][length-1] == 'A') ba++;
        else{
            if (s[i][0] == 'B') b_++;
            if (s[i][length-1] == 'A') _a++;
        }
        for (int j = 0; j < length - 1; j++){
            if (s[i].substr(j, 2) == "AB") ans++;
        }
    }
    int d = b_ - _a;
    if (d == 0 && ba != 0) ans += b_ + ba - 1;
    else if (d == 0 && ba == 0) ans += b_;
    else if (d > 0) ans += _a + ba;
    else if (d < 0) ans += b_ + ba;

    cout << ans << endl;
    return 0;
}