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 <bits/stdc++.h>
using namespace std;
 
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl "\n"
#define int long long

const int N = 1e5 + 5;

int n;
string s[N];

int get(string &x)
{
	int ans = 0;
	for(int i = 0; i + 1 < x.size(); i++)
		if(x[i] == 'A' && x[i + 1] == 'B')
			ans++;
	return ans;
}

int32_t main()
{
	IOS;
	cin >> n;
	int ans = 0;
	int endA = 0, startB = 0;
	bool check = 1;
	for(int i = 1; i <= n; i++)
	{
		cin >> s[i];
		ans += get(s[i]);
		endA += (s[i].back() == 'A');
		startB += (s[i][0] == 'B');
		if(s[i].back() != 'A' && s[i][0] != 'B')
			continue;
		bool cur = (s[i].back() == 'A');
		cur &= (s[i][0] == 'B');
		check &= cur;
	}
	ans += min({endA, startB});
	ans -= check;
	cout << ans;
	return 0;
}