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)

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

#define endl "\n"

//typedef
//------------------------------------------
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef vector<PII> VP;

//REPEAT
//------------------------------------------
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)

//container util
//------------------------------------------
#define pb push_back
#define paired make_pair
#define ALL(a) (a).begin(),(a).end()
#define PRINT(V) for (auto v : (V)) cout << v << " "
#define SORT(V) sort((V).begin(),(V).end())
#define RSORT(V) sort((V).rbegin(), (V).rend())

//constant
//------------------------------------------
const int MOD = 1000000007;
const int INF = 1061109567;
const double EPS = 1e-10;
const double PI  = acos(-1.0);
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};

//math
//------------------------------------------
int QP(int a,int b){int ans=1;do{if(b&1)ans=1ll*ans*a%MOD;a=1ll*a*a%MOD;}while(b>>=1);return ans;}
int QP(int a,int b,int MOD){int ans=1;do{if(b&1)ans=1ll*ans*a%MOD;a=1ll*a*a%MOD;}while(b>>=1);return ans;}
int GCD(int a,int b){return b?GCD(b,a%b):a;}

//debug
//------------------------------------------
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;

//grobal variable
//------------------------------------------
int N,M;
VI Vec;

//動的計画法で役に立つ関数
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

int main() 
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    
    int N; cin >> N;

    int n_a=0, n_b=0, n_ab=0;

    REP(i, N)
    {
        string s;cin >> s;
        if (s.front() == 'B') n_b++;
        if (s.back() == 'A' ) n_a++;
        bool was_A = false;
        for (auto c : s)
        {
            if (was_A && c=='B') n_ab++;

            if (c=='A') was_A=true;
            else was_A=false;
        }
    }
    cout << n_ab+min(n_a, n_b) << endl;


  
    return 0;
}