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>
using namespace std;
using LL = long long;
using ULL = unsigned long long;
const double PI = acos(-1);
template<class T>constexpr T INF() { return ::std::numeric_limits<T>::max(); }
template<class T>constexpr T HINF() { return INF<T>() / 2; }
template <typename T_char>T_char TL(T_char cX) { return tolower(cX); };
template <typename T_char>T_char TU(T_char cX) { return toupper(cX); };
template<typename T>vector<T> make_v(size_t a){ return vector<T>(a); }
template<typename T,typename... Ts>auto make_v(size_t a, Ts... ts){ return vector<decltype(make_v<T>(ts...))>(a,make_v<T>(ts...)); }
template<typename T,typename V>typename enable_if<is_class<T>::value==0>::type fill_v(T &t, const V &v){ t = v; }
template<typename T,typename V>typename enable_if<is_class<T>::value!=0>::type fill_v(T &t, const V &v){ for(auto &e : t) fill_v(e, v); }
const int vy[] = { -1, -1, -1, 0, 1, 1, 1, 0 }, vx[] = { -1, 0, 1, 1, 1, 0, -1, -1 };
const int dx[4] = { 0,1,0,-1 }, dy[4] = { 1,0,-1,0 };
int popcnt(unsigned long long n) { int cnt = 0; for (int i = 0; i < 64; i++)if ((n >> i) & 1)cnt++; return cnt; }
int d_sum(LL n) { int ret = 0; while (n > 0) { ret += n % 10; n /= 10; }return ret; }
int d_cnt(LL n) { int ret = 0; while (n > 0) { ret++; n /= 10; }return ret; }
LL gcd(LL a, LL b) { if (b == 0)return a; return gcd(b, a%b); };
LL lcm(LL a, LL b) { LL g = gcd(a, b); return a / g*b; };
# define ALL(qpqpq)           (qpqpq).begin(),(qpqpq).end()
# define UNIQUE(wpwpw)        sort(ALL((wpwpw)));(wpwpw).erase(unique(ALL((wpwpw))),(wpwpw).end())
# define LOWER(epepe)         transform(ALL((epepe)),(epepe).begin(),TL<char>)
# define UPPER(rprpr)         transform(ALL((rprpr)),(rprpr).begin(),TU<char>)
# define FOR(i,tptpt,ypypy)   for(LL i=(tptpt);i<(ypypy);i++)
# define REP(i,upupu)         FOR(i,0,upupu)
# define INIT                 std::ios::sync_with_stdio(false);std::cin.tie(0)

int n;
string s[10101];
int p, q, r;
int ans;
int main(){
    cin >> n;
    REP(i, n)cin >> s[i];
    REP(i, n){
        for(int j = 0;j < (int)s[i].size() - 1;j++){
            if(s[i][j] == 'A' && s[i][j + 1] == 'B')ans++;
        }
    }
    REP(i, n){
        if(s[i][0] == 'B' && s[i][(int)s[i].size() - 1] == 'A')r++;
        else if(s[i][0] == 'B')p++;
        else if(s[i][(int)s[i].size() - 1] == 'A')q++;
    }
    if(min(p, q) >= r){
        cout << ans + r*2 + min(p, q) - r << endl;
    }else if(max(p, q) >= r){
        cout << ans + min(p, r) + min(q, r) << endl;
    }else{
        cout << ans + p + q + (r - max(p, q)) << endl;
    }
}