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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

using ll = long long;
using ull = unsigned long long;
using ld = long double;

#define MOD 1000000007LL

#define rep(i, n) for(ll (i) = 0LL;(i) < (ll)(n);(i)++)
#define rep2(i, s, e) for(ll (i) = (ll)(s);(i) < (ll)(e);(i)++)
#define repi(i, n) for(ll (i) = 0LL;(i) <= (ll)(n);(i)++)
#define repi2(i, s, e) for(ll (i) = (ll)(s);(i) <= (ll)(e);(i)++)
#define per(i, n) for(ll (i) = (ll)(n) - 1LL;(i) >= 0LL;(i)--)
#define per2(i, s, e) for(ll (i) = (ll)(s) - 1LL;(i) >= (ll)(e);(i)--)
#define peri(i, n) for(ll (i) = (ll)(n);(i) >= 0LL;(i)--)
#define peri2(i, s, e) for(ll (i) = (ll)(s);(i) >= (ll)(e);(i)--)
#define iter(i, it) for(auto &(i): (it))

template<typename T, typename U> ostream& operator<<(ostream &s, const pair<T, U> m) {
    cout << "(" << m.first << ", " << m.second << ")";
    return s;
}
template<typename T, typename U> ostream& operator<<(ostream &s, const map<T, U> m) {
    ll c = 0;
    cout << "{ ";
    iter(i, m) cout << i << (c++ == m.size() - 1 ? " " : ", ");
    cout << "}";
    return s;
}
template<typename T> ostream& operator<<(ostream &s, const vector<T> &v) {
    cout << "{ ";
    rep(i, v.size()) cout << v[i] << (i == v.size() - 1 ? " " : ", ");
    cout << "}";
    return s;
}
template<typename T> ostream& operator<<(ostream &s, const list<T> &v) {
    ll c = 0;
    cout << "{ ";
    iter(i, v) cout << i << (c++ == v.size() - 1 ? " " : ", ");
    cout << "}";
    return s;
}

int main(void) {
    ll N;
    ll ab = 0;
    ll a = 0, b = 0, a2 = 0;
    cin >> N;
    rep(i, N) {
        string s;
        cin >> s;
        if(s.front() == 'B' && s.back() == 'A') a2++;
        else if(s.front() == 'B') b++;
        else if(s.back() == 'A') a++;
        rep2(i, 1, s.size()) {
            if(s[i - 1] == 'A' && s[i] == 'B') ab++;
        }
    }
    ll ans = ab + min(a, b);
    if(a + b == 0) ans += a2 - 1;
    else ans += a2;
    cout << ans << endl;
    return 0;
}