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++)

using namespace std;

int main() {
    int n,cnt = 0, a_last = 0, b_first = 0, u = 0;
    cin >> n;
    string s;
    rep(i, n){
        cin >> s;
        rep(j, s.length() - 1){
            if(s[j] == 'A' && s[j+1] == 'B'){
                cnt++;
            }
        }
        if(s[s.length() - 1] == 'A' && s[0] == 'B'){
            u++;
            continue;
        }
        if(s[s.length() - 1] == 'A') a_last++;
        if(s[0] == 'B') b_first++;
    }
    int diff = abs(a_last - b_first);
    if(diff >= u) cout << cnt + min(a_last, b_first) + u << endl;
    if(diff < u) cout << cnt + (u - diff) / 2 << endl;
}