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 <algorithm>
#include <cctype>
#include <cstdio>
#include <iostream>
#include <locale>
#include <numeric>
#include <string>
#include <vector>
 
using namespace std;

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

    int count=0;


    for(int i=0;i<n;i++){
        cin>>kp;
        s.push_back(kp);
    }

    for(int i=0;i<n;i++){
        for(int j=0;j<s[i].size();j++){
            if(s[i][j]=='A'){
                j++;
                if(s[i][j]=='B')count++;
            }
        }
    }
    // cout<<count<<endl;

    int ab=0;
    int beginB=0;
    int endA=0;

    for(int i=0;i<n;i++){
        if(s[i].front()=='B'&&s[i].back()=='A')ab++;
        else if(s[i].front()=='B')beginB++;
        else if(s[i].back()=='A')endA++;
    }
    // cout<<beginB<<" "<<endA<<" "<<ab;
    if(ab==0)cout<<count+min(beginB,endA);
    else if(endA!=0||beginB!=0)cout<<count+ab+min(beginB,endA);
    else cout<<count+ab-1;
    cout<<endl;
}