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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <string>

using namespace std;

int main()
{
	int N;
	cin >> N;

	unsigned long long counter = 0;
	unsigned long long Acnt = 0, Bcnt = 0, ABcnt = 0;
	vector<string> v( N );
	for ( int i = 0; i < N; i++ )
	{
		cin >> v[i];

		bool Afind = false;
		string::size_type index, beforeIndex = 0;
		while ( ( index = v[i].find( "AB", beforeIndex ) ) != string::npos )
		{
			counter++;
			beforeIndex = index + 1;
		}

		if ( v[i].back() == 'A' )
		{
			Afind = true;
			++Acnt;
		}
		if ( v[i][0] == 'B' )
		{
			++Bcnt;
			if ( Afind )
			{
				ABcnt++;
			}
		}
	}

	// AもBも含む文字列の方を先に考える
	if ( ABcnt > 0 )
	{
		counter += ( ABcnt - 1 ); // ループさせることはできないので、1パターン減る
	}

	// 出現頻度が少ない方に合わせる
	if ( Acnt > Bcnt )
	{
		counter += Bcnt;
	}
	else if ( Bcnt > Acnt )
	{
		counter += Acnt;
	}
	else
	{
		if ( Acnt == ABcnt )
		{
			if ( Acnt == 0 )
			{
				// 候補なし
			}
			else
			{
				counter += ( ABcnt - 1 );
			}
		}
	}


	cout << counter << endl;

	return ( 0 );
}