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 REP(i, n) for (int i = 0; i < (n); i++)
using namespace std;

int main() {
	cin.tie(0)->sync_with_stdio(false);

	int n; cin >> n;
	int cntAB = 0;
	int cntA = 0;
	int cntB = 0;
	int cntBA = 0;
	string find_word("AB");
	REP(i, n) {
		string s; cin >> s;
		auto pos = s.find(find_word);
		while (pos != string::npos) {
			cntAB++;
			pos = s.find(find_word, pos + find_word.length());
		}
		if (s.front() == 'B') cntB++;
		if (s.back() == 'A') cntA++;
		if (s.front() == 'B' && s.back() == 'A') cntBA++;
	}

	cout << cntAB + min(cntB, cntA) - cntBA / 2 << '\n';
	return 0;
}