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.Text.RegularExpressions;
using System.Threading.Tasks;
using static System.Math;

namespace AtCorder
{
    class Program
    {
        public static void Main(string[] args)
        {
            var n = int.Parse(Console.ReadLine());
            var a = Enumerable
                .Range(0, n)
                .Select(x => Console.ReadLine())
                .ToArray();

            var regex = new Regex("AB");
            var numOfBegin = 0;
            var numOfEnd = 0;
            var numOfBoth = 0;
            var count = 0;
            foreach (var l in a)
            {
                count += regex.Matches(l).Count;
                var begin = l.StartsWith("B");
                var end = l.EndsWith("A");

                if (begin && end)
                {
                    numOfBoth++;
                }
                else if (begin)
                {
                    numOfBegin++;
                }
                else if (end)
                {
                    numOfEnd++;
                }
            }

            bool hasA = false;
            bool hasB = false;
            if (numOfBoth > 0)
            {
                count += numOfBoth - 1;
                hasA = true;
                hasB = true;
            }

            if (hasA && numOfEnd > 0)
            {
                count++;
                numOfEnd--;
            }

            if (hasB && numOfBegin > 0)
            {
                count++;
                numOfBegin--;
            }

            count += Min(numOfBegin, numOfEnd);

            Console.WriteLine(count);
        }
    }
}