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 chmax(x, y) x = max(x, y)
#define chmin(x, y) x = min(x, y)
using namespace std;
using ll = long long;

int calc(int ab, int b, int a) {
    if (b == 0 && a == 0) {
        return ab - 1;
    }
    return ab + min(b, a);
}

int main() {
    int N;
    cin >> N;
    vector<string> s(N);

    map<int, int> mp;
    int ans = 0;
    rep(i, N) {
        string s;
        cin >> s;
        rep(i, s.size() - 1) {
            if (s[i] == 'A' && s[i + 1] == 'B') ans++;
        }
        bool beginB = s[0] == 'B';
        bool endA = s[s.size() - 1] == 'A';
        if (beginB && endA) {
            mp[0]++;
        } else if (beginB) {
            mp[1]++;
        } else if (endA) {
            mp[2]++;
        }
    }

    ans += calc(mp[0], mp[1], mp[2]);
    cout << ans << endl;
}