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>
#define _GLIBCXX_DEBUG
#define rep(i, n) for(int i = 0; i < (n); ++i)
#define rep1(i, n) for(int i = 1; i < (n); ++i)
using namespace std;
typedef long long ll;

int countAB(string s){
    int ret = 0;
    rep(i, s.size()-1){
        if(s[i]=='A' && s[i+1]=='B') ++ret;
    }
    return ret;
}

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);

    int n;
    cin >> n;

    vector<string> s(n);
    int res = 0;
    int frontB = 0;
    int backA = 0;
    int both = 0;
    rep(i, n){
        cin >> s[i];
        res += countAB(s[i]);
        if(s[i].front() == 'B' && s[i].back() == 'A') ++both;
        if(s[i].front() == 'B') ++frontB;
        if(s[i].back() == 'A') ++backA;
    }
    res += min(frontB, backA);
    if(frontB==backA && backA==both) --res;
    
    cout << res  << endl;

    return 0;
}