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>
#include <iomanip>
#include <math.h>
#include <unistd.h>
#include <stdio.h>
#define all(a) (a).begin(),(a).end()
#define vecin(a) rep(i,a.size())cin >> a[i]
#define vecout(a) rep(i,a.size()){cout << a[i];cout << (i == a.size() - 1 ? "\n":" ");}
#define rep(i,x) for(int i = 0;i<x;i++)
#define REP(i,x) for(int i = 1;i<=x;i++)
#define mp make_pair
using ll = long long;
using ld = long double;
using namespace std;
using dou = double;
const ll inf = 2147483647;
const ll INF = 1LL << 60;
const dou pi = 3.14159265358;
//const ll mod = 1000000007LL;
const ll mod = 998244353LL;
typedef pair<ll,ll> P;
using graph = vector<vector<ll>>;
template<class T> inline bool chmax(T& a, T b){if(a < b){a = b; return true;}return false;}
template<class T> inline bool chmin(T& a, T b){if(a > b){a = b; return true;}return false;}
template<class T> inline bool change(T a,T b){if(a > b){swap(a,b);return true;}return false;}
ll gcd(ll a,ll b){
    if(a < b)swap(a,b);
    if(a % b == 0)return b;
    else return gcd(b,a%b);
}
ll lcm(ll a,ll b){
    return a / gcd(a,b) * b;
}

//素数判定O(sqrt(N))
bool isp(ll n){
    bool res = true;
    if(n == 1)return false;
    else{
        for(ll i = 2;i * i <= n;i++){
            if(n % i == 0){
                res = false;
                break;
            }
        }
        return res;
    }
}
const ll cmax = 1000000;
ll fac[cmax], finv[cmax], inv[cmax];

// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (ll i = 2; i < cmax; i++){
        fac[i] = fac[i - 1] * i % mod;
        inv[i] = mod - inv[mod%i] * (mod / i) % mod;
        finv[i] = finv[i - 1] * inv[i] % mod;
    }
}

// 二項係数計算
ll nCk(ll n, ll k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % mod) % mod;
}
//nのm乗をMODで割ったあまりO(logm)
ll modpow(ll n,ll m,ll MOD){
    if(m == 0)return 1;
    if(m < 0)return -1;
    ll res = 1;
    while(m){
        if(m & 1)res = (res * n) % MOD;
        m >>= 1;
        n *= n;
        n %= MOD;
    }
    return res;
}
ll mypow(ll n,ll m){
    if(m == 0)return 1;
    if(m < 0)return -1;
    ll res = 1;
    while(m){
        if(m & 1)res = (res * n);
        m >>= 1;
        n *= n;
    }
    return res;
}
//printf("%.10f\n", N);
/*  vector<ll> vec = {1,2,5,6,8,9,10};
    cout << binary_search(all(vec),5) << endl; -> true*/
/*
    vector<ll> vec = {1,2,5,6,8,9,10};
    auto n = upper_bound(all(vec),4);
    cout << *n << endl; -> 5*/

//    cout << fixed << setprecision(15);


int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    ll N;
    cin >> N;
    vector<string> vec(N);
    vecin(vec);
    vector<ll> naka(N);//...AB...
    ll A = 0,B = 0,both = 0;//....A/B....
    rep(i,N){
        ll s = vec[i].size();
        rep(j,s - 1){
            if(vec[i][j] == 'A' && vec[i][j + 1] == 'B'){
                j++;
                naka[i]++;
            }
        }
        if(vec[i][s - 1] == 'A' && vec[i][0] == 'B')both++;
        else if(vec[i][0] == 'B')B++;
        else if(vec[i][s - 1] == 'A')A++;
    }
    ll ans = 0;
    rep(i,N){
        ans += naka[i];
    }
    //vecout(naka);
    //cout << A << "\n" << B << "\n" << both << "\n";
    ll n = min(min(both,A),B);
    both -= n;A -= n;B -= n;
    ans += n * 2;
    if(both == 0){
        ans += min(A,B);
    }
    else{
        ans += both - 1;
        if(A || B)ans++;
    }
    cout << ans << endl;
}