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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.IO;
using static System.Console;
using static System.Math;

namespace AtCodeeeer
{
    class AtCOOOOOOOOOOOder
    {
        static void Main(string[] args)
        {
            int N = int.Parse(ReadLine());
            int BほにゃA = 0;int BほにゃnotA = 0;int notBほにゃA = 0;
            int ans = 0;
            for (int i = 0; i < N; i++)
            {
                string s = ReadLine();

                if (s.StartsWith("B")&&s.EndsWith("A"))
                {
                    BほにゃA++;
                }
                if (s.StartsWith("B")&& !s.EndsWith("A"))
                {
                    BほにゃnotA++;
                }
                if (!s.StartsWith("B") && s.EndsWith("A"))
                {
                    notBほにゃA++;
                }
                
                for (int j = 0; j < s.Length-1; j++)
                {
                    if (s.Substring(j,2) == "AB")
                    {
                        ans++;
                    }
                }
            }
            if (BほにゃA == 0)
            {
                WriteLine(ans + Min(BほにゃnotA, notBほにゃA));
            }
            else if (BほにゃnotA + notBほにゃA == 0)
            {
                WriteLine(ans + BほにゃA - 1);
            }
            else
            {
                WriteLine(ans + BほにゃA + Min(BほにゃnotA, notBほにゃA));
            }
        }
    }
}