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: Accepted)

#include <iostream>
#include <vector>
#include <string>
#include <numeric>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
#include <cmath>

using namespace std;
using ll = long long;

int gcd(int a, int b) { return (b == 0) ? a : gcd(b, a % b); }

#define CIN(T, x) T x; cin >> x;
#define CINS(T, x, N) vector<T> x(N); for(int i = 0; i < (N); i++) { cin >> x[i]; }
#define SHOW(x) for(int i = 0; i < x.size(); i++) { cout << x[i] << " "; } cout << "\n";
#define SHOW2(x) for(int j = 0; j < x.size(); j++) { SHOW(x[j]); } cout << "\n";


int main() {

    CIN(int, N);
    CINS(string, s, N);

    int countAB = 0;
    int countAandB = 0;
    int countAonly = 0;
    int countBonly = 0;

    for (int i = 0; i < N; i++) {
        string si = s[i];

        if (si[0] == 'B' && si[si.size()-1] == 'A') {
            countAandB++;
        } else if (si[si.size()-1] == 'A') {
            countAonly++;
        } else if (si[0] == 'B') {
            countBonly++;
        }

        for (int j = 0; j < s[i].size()-1; j++) {
            if ( s[i][j] == 'A' && s[i][j+1] == 'B' ) {
                countAB++;
            }
        }
    }

    int ans = countAB;

    ans += min(countAonly + countAandB, countBonly + countAandB);
    if ( countAonly == 0 && countBonly == 0 && countAandB > 0 ) {
        ans--;
    }

    cout << ans << "\n";
    return 0;
}