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<iostream>
#include<vector>
#include<algorithm>
#include<string>
using namespace std;
#define ll long long

int main(void){
    int N;
    int ans=0;
    cin>>N;
    vector<string> letter(N);
    vector<char> head(N);
    vector<char> back(N);
    for(int i=0;i<N;i++) {
        cin>>letter.at(i);
        for(int j=0;j<letter.at(i).size()-1;j++) if(letter.at(i).at(j)=='A' && letter.at(i).at(j+1)=='B') ans++;
        head.at(i)=letter.at(i).at(0);
        back.at(i)=letter.at(i).at(letter.at(i).size()-1);
    }


    int BA=0,BnotA=0,notBA=0;
    for(int i=0;i<N;i++) {
        if(head.at(i)=='B' && back.at(i)=='A') BA++;
        else if(head.at(i)=='B') BnotA++;
        else if(back.at(i)=='A') notBA++;
    }
    ans+=BA+min(BnotA,notBA);
    if(BnotA==0 && notBA==0) ans--;
    
    cout<<ans<<endl;
    return 0;
}