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)

#define _USE_MATH_DEFINES

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>

using namespace std;

typedef pair<long long int, long long int> P;

long long int INF = 1e18;
long long int MOD = 1e9 + 7;

int main(){
    int N;
    cin >> N;
    long long int ans = 0;
    string s;
    int cnt1 = 0, cnt2 = 0;
    bool flag = true;
    for(int i = 0; i < N; i++){
        cin >> s;
        for(int j = 0; j < s.size() - 1; j++){
            if(s[j] == 'A' && s[j + 1] == 'B'){
                ans += 1;
            }
        }
        if(s[0] == 'B'){
            cnt1 += 1;
        }
        if(s[s.size() - 1] == 'A'){
            cnt2 += 1;
        }
        if(s[s.size() - 1] == 'A' && s[0] != 'B'){
            flag = false;
        }
    }
    ans += min(cnt1, cnt2);
    if(flag){
        ans -= 1;
    }
    cout << ans << endl;
    return 0;
}